GEOMETRICAL OPTICS
EXERCISES- Q61 to Q70
61. A diverging lens of focal length 20 cm and a converging mirror of focal length 10 cm are placed coaxially at a separation of 5 cm. Where should an object be placed so that a real image is formed at the object itself.
ANSWER: First Method
Let the object be at x cm from the lens. Taking the origin at the optical center of the lens.
u = -x cm, f = -20 cm
1/v - 1/(-x) = 1/(-20)
→1/v = -1/20 - 1/x = -(x+20)/20x
→v =-20x/(x+20) -------------------(i)
So this virtual image is towards the object but it is the object for the mirror. Now taking the origin at the center of the mirror. Object distance
u = -5 -20x/(x+20) =-(5x+100+20x)/(x+20)
=-(25x+100)/(x+20)
f =-10 cm
From the mirror formula
1/v+1/u =1/f
→1/v-(x+20)/(25x+100) = -1/10
→1/v = (x+20)/(25x+100) - 1/10
→1/v =(10x+200-25x-100)/10(25x+100)
→1/v = (10-1.5x)/(25x+100)
→v = -(25x+100)/(1.5x-10)
Diagram for Q-61
This image distance is from mirror center C and in front of the mirror because the sign is negative. Again to form the image at the object the rays will go for second refraction through the lens and this image will be object for the lens. Assuming that the image is between the lens and the mirror (now origin will be taken at O again),
Now u =-[5-(25x+100)/(1.5x-10)]
{The value of v is taken without sign i.e. numerical value}
→u =-(7.5x-50-25x-100)/(1.5x-10)
=(17.5x+150)/(1.5x-10)
v = x
f =-20 cm
From the lens formula
1/v -1/u = 1/f
→1/x -(1.5x-10)/(17.5x+150) = -1/20
→(17.5x+150-1.5x²+10x)/x(17.5x+150) =-1/20
→20(-1.5x²+27.5x+150) =-17.5x²-150x
→-30x²+17.5x²+550x+150x+3000 =0
→-12.5x²+700x+3000=0
→x²-56x-240 =0
→x ={56士√(56²+4*240)}/2
={56士64}/2
=120/2 or -8/2
=60 or -4
In the given condition we neglect the negative value because it will be between the lens and the mirror and not real. So we have x = 60 cm i.e. the object should be placed 60 cm from the lens further away from the mirror.
Alternately
If the rays after reflection from the mirror trace back the same path, then the real image will be on the object itself. This will happen when the rays fall on the mirror perpendicularly. The rays will fall perpendicularly on the mirror if they will appear to come from the center of curvature.
Diagram for Q-61
Thus in the figure below, CR = radius of curvature =2f =2*10 =20 cm.
So, OR =20-5 =15 cm = image distance for the first refraction. But from (i) it is equal to 20x/(x+20). Equating these two,
20x/(x+20) =15
→20x =15x+300
→5x =300
→x =300/5 =60 cm.
62. A converging lens of focal length 12 cm and a diverging mirror of focal length 7.5 cm are placed at 5.0 cm apart with their principal axis coinciding. Where should an object be placed so that its image falls on itself?
ANSWER: For the lens f = 12 cm, assume the object is at distance x cm from the lens, so u = -x. From the lens formula
1/v-1/u =1/f
→1/v + 1/x =1/12
→1/v = 1/12 - 1/x =(x-12)/12x
→v = 12x/(x-12)
This image is the object for the mirror. If the rays after the first refraction fall perpendicularly on the mirror, then they will trace back and form the image on the object itself. This condition will be fulfilled if this image is formed on the center of curvature of the mirror R.
Diagram for Q-62
Thus OR = v =12x/(x-12)
→OC+CR =12x/(x-12)
{CR =radius of curvature =2f}
→5 + 2f =12x/(x-12)
→5+2*7.5 =12x/(x-12)
{for mirror, f =7.5 cm}
→12x/(x-12) =20
→12x =20x-240
→8x = 240
→x =240/8 =30 cm
So the object should be placed 30 cm from the lens further away from the mirror.
63. A converging lens and a diverging mirror are placed at a separation of 15 cm. The focal length of the lens is 25 cm and that of the mirror is 40 cm. Where should a point source be placed between the lens and the mirror so that the light, after getting reflected by the mirror and then getting transmitted by the lens, comes out parallel to the principal axis?
ANSWER: The rays come out parallel to the principal axis after refraction from a lens if the object is at its focus F. So in this case after reflection from the mirror the virtual image will be at F.
Diagram for Q-63
Suppose the object is placed at P at a distance x from the lens, OP =x. For the mirror, u =-(15-X), f =40 cm. From the mirror formula
1/v + 1/u = 1/f
→1/v-1/(15-x) =1/40
→1/v =1/(15-x) + 1/40 =(40+15-x)/{40*(15-x)}
→1/v =(55-x)/{40(15-x)}
→v = (600-40x)/(55-x) =CF =OF-OC =25-15 =10
{OF = focal length of the lens =25 cm}
→600-40x =550-10x
→30x =50
→x = 50/30 =1.67 cm
So the object should be kept at 1.67 cm from the lens.
64. A converging lens of focal length 15 cm and a converging mirror of focal length 10 cm are placed 50 cm apart with a common principal axis. A point source is placed in between the lens and the mirror at a distance of 40 cm from the lens. Find the locations of the two images formed.
ANSWER: One image will be formed after the refraction from the lens while the other will be formed first from the reflection of the mirror and then refraction from the lens.
From the given data the point source is on the focus of the converging mirror. After reflection from the mirror, the rays will go parallel and fall on the lens. These parallel rays will make an image at the focus of the lens i.e. at 15 cm from the lens on the other side.
For the image formed only by the lens, u = -40 cm, f= 15 cm, v =? From the lens formula
1/v - 1/u = 1/f
→1/v = 1/15 -1/40 =(8-3) /120 =5/120 =1/24
→v = 24 cm on the other side of the lens.
65. Consider the situation described in the previous problem. Where should a point source be placed on the principal axis so that the two images form at the same place?
ANSWER: Suppose such place is at a distance of x cm from the lens between them. After direct refraction from the lens, let the image be at y cm from the lens. In the second case where the first reflection from the mirror takes place if the image is formed at the object then after refraction from the lens again the final image will be at the place where direct refraction image was formed i.e. at y cm from the lens. Now the condition that the image formed by the mirror is at the object itself is that the object is at the radius of the curvature of the mirror. i.e. at 2f =2*10 cm =20 cm from the mirror. So the object should be placed at 50-20 =30cm from the lens towards the mirror.
66. A converging lens of focal length 15 cm and a converging mirror of focal length 10 cm are placed 50 cm apart. If a pin of length 2.0 cm is placed 30 cm from the lens farther away from the mirror, where will the final image form and what will be the size of the final image?
ANSWER: For the lens f=15 cm. Hence R=2f=30 cm.
The pin is placed at R of the converging lens. We know that for a converging lens if the object is placed at R, the inverted, real and same size image forms at R on the other side. In this case, the object AR will have an image R'D at 30 cm from the lens. It will be of the same size but inverted. This image will act as an object for the mirror. The focal length of the mirror f = 10 cm, Radius of curvature =CR' =2*10 = 20 cm. Since the distance between lens and mirror =OC= 50 cm, the object is 50-30 =20 cm from the mirror i.e. it is at the radius of curvature of the converging mirror. Its real image of the same size will be formed at the same place but inverted ER'. See the figure below:-
Diagram for Q-66
For the second refraction through the lens, this image is an object placed at the radius of curvature. So, again the same size, a real and inverted image on the other side of the lens at the 2f=30 cm distance will be formed i.e. at the real object itself.
67. A point object is placed on the principal axis of a convex lens (f = 15 cm) at a distance of 30 cm from it. A glass plate (µ = 1.50) of thickness 1 cm is placed on the other side of the lens perpendicular to the axis. Locate the image of the point object.
ANSWER: Since the object is placed at the 2f =2*15=30 cm distance from the lens, hence the image will be formed at 2f = 30 cm distance from the lens on the other side. If the 1 cm thick glass is placed within this distance there will be a shift in the image =(1-1/µ)*t =(1-1/1.5)*1 cm
=0.5/1.5 cm =0.33 cm.
Hence the image will be at 30+0.33 =30.33 cm from the lens.
68. A convex lens of focal length 20 cm and a concave lens of focal length 10 cm are placed 10 cm apart with their principal axes coinciding. A beam of light traveling parallel to the principal axis and having a beam diameter 5.0 mm, is incident on the combination. Show that the emergent beam is parallel to the incident one. Find the beam diameter of the emergent beam.
ANSWER: When the beam of light is incident from the side of the concave lens:-
Diagram for Q-68
Due to the position of the lenses, F is the common focus. PA is the topmost ray parallel to the principal axis. After refraction from the concave lens, it will appear to be coming from F and go along AB, i.e. the virtual image will be formed at the focus F. It can also be found out by the lens formula,
1/v +1/∞ =-1/10 {u =-∞, f =-10 cm}
→1/v =-1/10
→v = -10 cm =f
This virtual image will act as an object for the convex lens. Now F is also the focus of the convex lens. The rays coming from the focus of a convex lens go parallel to the principal axis. Hence the ray AB will go along BD after refraction from the convex lens, so the final image will be at infinity. It can also be found out from the lens formula,
1/v -1/(-20)=1/20 {u =-20 cm, f =20 cm}
→1/v =1/20-1/20 =0
→v =∞.
To get the beam diameter we compare triangles FAO' and FBO. Both are similar triangles, hence the corresponding sides will be proportionate.
BO/AO' =OF/OF'
→BO =AO'*OF/OF' =(5 mm)*(20 cm/10 cm)
→BO =5 mm*2 =10 mm =1 cm.
When the beam is incident from the side of the convex lens:-
Diagram for Q-68
The topmost ray PA which is parallel to the principal axis will form an image at the Focus F after refraction from the convex lens. This image will act as an object for the concave lens. An object at the focus makes the image at infinity, we can check it from the lens formula. u = O'F =10 cm, f =-10 cm, v =?
1/v - 1/10 =-1/10
→1/v =0
→v =∞, so the rays emerge parallel to the principal axis after refraction from the concave lens. For the beam diameter of the emergent beam, we find that the triangles AOF and BO'F are similar. Hence the corresponding sides will be proportionate.
BO'/AO =O'F/OF
→BO' =AO*O'F/OF =(5 mm)*(10 cm/20 cm)
→BO' =5 mm*(1/2) =2.5 mm.
69. A diverging lens of focal length 20 cm and a converging lens of focal length 30 cm are placed 15 cm apart with their principal axes coinciding. Where should an object be placed on the principal axis so that its image is formed at infinity?
ANSWER: For the diverging lens f₁ = -20 cm,
For the converging lens f₂ = 30 cm,
the distance between the lenses d =15 cm, hence the equivalent focal length of the combined lens, F is given as
1/F =1/f₁+1/f₂-d/f₁f₂
→1/F =-1/20+1/30-15/(-20)*30
→1/F =1/30-1/20+1/40
→1/F =(4-6+3)/120 =1/120
→F =120 cm
The positive sign shows that the equivalent lens is a converging lens. This focal length is the distance from the point P on the principal axis where the equivalent single lens is to be kept. The distance of the point P behind the converging lens
PO' =d*F/f₁ =15*120/(-20) =-90 cm. Negative sign shows that this distance is in the front of the converging lens.
Diagram for Q-69
If the object is kept at the focus of a converging lens, the image is formed at infinity on the other side. If the object is placed on the focus F₂ on the side of the convex lens, then PF₂ = 120 cm. The distance of the object from the converging lens
O'F₂ =O'P+PF₂ =90+120 =210 cm.
For the object placed on the other side, the distance of point P behind the diverging lens
PO =d*F/f₂ =15*120/30 =60 cm.
Diagram for Q-69
If the object is placed on the focus F₁ on the side of the diverging lens, then PF₁ =120 cm. The distance of the object from the diverging lens
OF₁ = PF₁-PO =120-60 =60 cm
70. A 5 mm high pin is placed at a distance of 15 cm from a convex lens of focal length 10 cm. A second lens of focal length 5 cm is placed 40 cm from the first lens and 55 cm from the pin. Find (a) the position of the final image, (b) its nature and (c) its size.
ANSWER: u = -15 cm, f =10 cm, v =?
1/v - 1/(-15) =1/10
→1/v =1/10-1/15 =5/150 =1/30
→v =30 cm
The image is formed on the other side of the convex lens. On this side, another lens at 40 cm from the first lens is kept. This image is now the object for the second lens. Object distance u =40 cm - 30 cm = 10 cm. Since the focal length of this lens is f = 5 cm, hence the object is placed at 2f =10 cm. For a convex lens, an object placed at 2f distance forms an image at the 2f distance on the other side of the lens.
(a) Hence the final image is at 2f = 10 cm from the second lens further away.
(b) Nature of the image:-
For the first lens, magnification m =v/u =hᵢ/hₒ
→hᵢ/hₒ =30/(-15) =-2
→hᵢ =-2hₒ =-2*5 mm =-10 mm
So the image is real and inverted.
For the second lens,
m =hᵢ/hₒ=v/u =2f/(-2f) =-1
→hᵢ = -hₒ =-10 mm
So the image is real and inverted and of the same size. Thus the first lens inverts the image and the second lens makes the final image erect by inverting again. So the final image is real and erect.
(c) Size of the image:-
As we saw in (b), the first lens makes the image size =10 cm and the second lens also makes the same size of the image, i.e. =10 cm.
61. A diverging lens of focal length 20 cm and a converging mirror of focal length 10 cm are placed coaxially at a separation of 5 cm. Where should an object be placed so that a real image is formed at the object itself.
ANSWER: First Method
Let the object be at x cm from the lens. Taking the origin at the optical center of the lens.
u = -x cm, f = -20 cm
1/v - 1/(-x) = 1/(-20)
→1/v = -1/20 - 1/x = -(x+20)/20x
→v =-20x/(x+20) -------------------(i)
So this virtual image is towards the object but it is the object for the mirror. Now taking the origin at the center of the mirror. Object distance
u = -5 -20x/(x+20) =-(5x+100+20x)/(x+20)
=-(25x+100)/(x+20)
f =-10 cm
From the mirror formula
1/v+1/u =1/f
→1/v-(x+20)/(25x+100) = -1/10
→1/v = (x+20)/(25x+100) - 1/10
→1/v =(10x+200-25x-100)/10(25x+100)
→1/v = (10-1.5x)/(25x+100)
→v = -(25x+100)/(1.5x-10)
This image distance is from mirror center C and in front of the mirror because the sign is negative. Again to form the image at the object the rays will go for second refraction through the lens and this image will be object for the lens. Assuming that the image is between the lens and the mirror (now origin will be taken at O again),
Now u =-[5-(25x+100)/(1.5x-10)]
{The value of v is taken without sign i.e. numerical value}
→u =-(7.5x-50-25x-100)/(1.5x-10)
=(17.5x+150)/(1.5x-10)
v = x
f =-20 cm
From the lens formula
1/v -1/u = 1/f
→1/x -(1.5x-10)/(17.5x+150) = -1/20
→(17.5x+150-1.5x²+10x)/x(17.5x+150) =-1/20
→20(-1.5x²+27.5x+150) =-17.5x²-150x
→-30x²+17.5x²+550x+150x+3000 =0
→-12.5x²+700x+3000=0
→x²-56x-240 =0
→x ={56士√(56²+4*240)}/2
={56士64}/2
=120/2 or -8/2
=60 or -4
In the given condition we neglect the negative value because it will be between the lens and the mirror and not real. So we have x = 60 cm i.e. the object should be placed 60 cm from the lens further away from the mirror.
Alternately
If the rays after reflection from the mirror trace back the same path, then the real image will be on the object itself. This will happen when the rays fall on the mirror perpendicularly. The rays will fall perpendicularly on the mirror if they will appear to come from the center of curvature.
Thus in the figure below, CR = radius of curvature =2f =2*10 =20 cm.
So, OR =20-5 =15 cm = image distance for the first refraction. But from (i) it is equal to 20x/(x+20). Equating these two,
20x/(x+20) =15
→20x =15x+300
→5x =300
→x =300/5 =60 cm.
62. A converging lens of focal length 12 cm and a diverging mirror of focal length 7.5 cm are placed at 5.0 cm apart with their principal axis coinciding. Where should an object be placed so that its image falls on itself?
ANSWER: For the lens f = 12 cm, assume the object is at distance x cm from the lens, so u = -x. From the lens formula
1/v-1/u =1/f
→1/v + 1/x =1/12
→1/v = 1/12 - 1/x =(x-12)/12x
→v = 12x/(x-12)
This image is the object for the mirror. If the rays after the first refraction fall perpendicularly on the mirror, then they will trace back and form the image on the object itself. This condition will be fulfilled if this image is formed on the center of curvature of the mirror R.
Thus OR = v =12x/(x-12)
→OC+CR =12x/(x-12)
{CR =radius of curvature =2f}
→5 + 2f =12x/(x-12)
→5+2*7.5 =12x/(x-12)
{for mirror, f =7.5 cm}
→12x/(x-12) =20
→12x =20x-240
→8x = 240
→x =240/8 =30 cm
So the object should be placed 30 cm from the lens further away from the mirror.
63. A converging lens and a diverging mirror are placed at a separation of 15 cm. The focal length of the lens is 25 cm and that of the mirror is 40 cm. Where should a point source be placed between the lens and the mirror so that the light, after getting reflected by the mirror and then getting transmitted by the lens, comes out parallel to the principal axis?
ANSWER: The rays come out parallel to the principal axis after refraction from a lens if the object is at its focus F. So in this case after reflection from the mirror the virtual image will be at F.
Suppose the object is placed at P at a distance x from the lens, OP =x. For the mirror, u =-(15-X), f =40 cm. From the mirror formula
1/v + 1/u = 1/f
→1/v-1/(15-x) =1/40
→1/v =1/(15-x) + 1/40 =(40+15-x)/{40*(15-x)}
→1/v =(55-x)/{40(15-x)}
→v = (600-40x)/(55-x) =CF =OF-OC =25-15 =10
{OF = focal length of the lens =25 cm}
→600-40x =550-10x
→30x =50
→x = 50/30 =1.67 cm
So the object should be kept at 1.67 cm from the lens.
64. A converging lens of focal length 15 cm and a converging mirror of focal length 10 cm are placed 50 cm apart with a common principal axis. A point source is placed in between the lens and the mirror at a distance of 40 cm from the lens. Find the locations of the two images formed.
ANSWER: One image will be formed after the refraction from the lens while the other will be formed first from the reflection of the mirror and then refraction from the lens.
From the given data the point source is on the focus of the converging mirror. After reflection from the mirror, the rays will go parallel and fall on the lens. These parallel rays will make an image at the focus of the lens i.e. at 15 cm from the lens on the other side.
For the image formed only by the lens, u = -40 cm, f= 15 cm, v =? From the lens formula
1/v - 1/u = 1/f
→1/v = 1/15 -1/40 =(8-3) /120 =5/120 =1/24
→v = 24 cm on the other side of the lens.
65. Consider the situation described in the previous problem. Where should a point source be placed on the principal axis so that the two images form at the same place?
ANSWER: Suppose such place is at a distance of x cm from the lens between them. After direct refraction from the lens, let the image be at y cm from the lens. In the second case where the first reflection from the mirror takes place if the image is formed at the object then after refraction from the lens again the final image will be at the place where direct refraction image was formed i.e. at y cm from the lens. Now the condition that the image formed by the mirror is at the object itself is that the object is at the radius of the curvature of the mirror. i.e. at 2f =2*10 cm =20 cm from the mirror. So the object should be placed at 50-20 =30cm from the lens towards the mirror.
66. A converging lens of focal length 15 cm and a converging mirror of focal length 10 cm are placed 50 cm apart. If a pin of length 2.0 cm is placed 30 cm from the lens farther away from the mirror, where will the final image form and what will be the size of the final image?
ANSWER: For the lens f=15 cm. Hence R=2f=30 cm.
The pin is placed at R of the converging lens. We know that for a converging lens if the object is placed at R, the inverted, real and same size image forms at R on the other side. In this case, the object AR will have an image R'D at 30 cm from the lens. It will be of the same size but inverted. This image will act as an object for the mirror. The focal length of the mirror f = 10 cm, Radius of curvature =CR' =2*10 = 20 cm. Since the distance between lens and mirror =OC= 50 cm, the object is 50-30 =20 cm from the mirror i.e. it is at the radius of curvature of the converging mirror. Its real image of the same size will be formed at the same place but inverted ER'. See the figure below:-
For the second refraction through the lens, this image is an object placed at the radius of curvature. So, again the same size, a real and inverted image on the other side of the lens at the 2f=30 cm distance will be formed i.e. at the real object itself.
67. A point object is placed on the principal axis of a convex lens (f = 15 cm) at a distance of 30 cm from it. A glass plate (µ = 1.50) of thickness 1 cm is placed on the other side of the lens perpendicular to the axis. Locate the image of the point object.
ANSWER: Since the object is placed at the 2f =2*15=30 cm distance from the lens, hence the image will be formed at 2f = 30 cm distance from the lens on the other side. If the 1 cm thick glass is placed within this distance there will be a shift in the image =(1-1/µ)*t =(1-1/1.5)*1 cm
=0.5/1.5 cm =0.33 cm.
Hence the image will be at 30+0.33 =30.33 cm from the lens.
68. A convex lens of focal length 20 cm and a concave lens of focal length 10 cm are placed 10 cm apart with their principal axes coinciding. A beam of light traveling parallel to the principal axis and having a beam diameter 5.0 mm, is incident on the combination. Show that the emergent beam is parallel to the incident one. Find the beam diameter of the emergent beam.
ANSWER: When the beam of light is incident from the side of the concave lens:-
Due to the position of the lenses, F is the common focus. PA is the topmost ray parallel to the principal axis. After refraction from the concave lens, it will appear to be coming from F and go along AB, i.e. the virtual image will be formed at the focus F. It can also be found out by the lens formula,
1/v +1/∞ =-1/10 {u =-∞, f =-10 cm}
→1/v =-1/10
→v = -10 cm =f
This virtual image will act as an object for the convex lens. Now F is also the focus of the convex lens. The rays coming from the focus of a convex lens go parallel to the principal axis. Hence the ray AB will go along BD after refraction from the convex lens, so the final image will be at infinity. It can also be found out from the lens formula,
1/v -1/(-20)=1/20 {u =-20 cm, f =20 cm}
→1/v =1/20-1/20 =0
→v =∞.
To get the beam diameter we compare triangles FAO' and FBO. Both are similar triangles, hence the corresponding sides will be proportionate.
BO/AO' =OF/OF'
→BO =AO'*OF/OF' =(5 mm)*(20 cm/10 cm)
→BO =5 mm*2 =10 mm =1 cm.
When the beam is incident from the side of the convex lens:-
The topmost ray PA which is parallel to the principal axis will form an image at the Focus F after refraction from the convex lens. This image will act as an object for the concave lens. An object at the focus makes the image at infinity, we can check it from the lens formula. u = O'F =10 cm, f =-10 cm, v =?
1/v - 1/10 =-1/10
→1/v =0
→v =∞, so the rays emerge parallel to the principal axis after refraction from the concave lens. For the beam diameter of the emergent beam, we find that the triangles AOF and BO'F are similar. Hence the corresponding sides will be proportionate.
BO'/AO =O'F/OF
→BO' =AO*O'F/OF =(5 mm)*(10 cm/20 cm)
→BO' =5 mm*(1/2) =2.5 mm.
69. A diverging lens of focal length 20 cm and a converging lens of focal length 30 cm are placed 15 cm apart with their principal axes coinciding. Where should an object be placed on the principal axis so that its image is formed at infinity?
ANSWER: For the diverging lens f₁ = -20 cm,
For the converging lens f₂ = 30 cm,
the distance between the lenses d =15 cm, hence the equivalent focal length of the combined lens, F is given as
1/F =1/f₁+1/f₂-d/f₁f₂
→1/F =-1/20+1/30-15/(-20)*30
→1/F =1/30-1/20+1/40
→1/F =(4-6+3)/120 =1/120
→F =120 cm
The positive sign shows that the equivalent lens is a converging lens. This focal length is the distance from the point P on the principal axis where the equivalent single lens is to be kept. The distance of the point P behind the converging lens
PO' =d*F/f₁ =15*120/(-20) =-90 cm. Negative sign shows that this distance is in the front of the converging lens.
If the object is kept at the focus of a converging lens, the image is formed at infinity on the other side. If the object is placed on the focus F₂ on the side of the convex lens, then PF₂ = 120 cm. The distance of the object from the converging lens
O'F₂ =O'P+PF₂ =90+120 =210 cm.
For the object placed on the other side, the distance of point P behind the diverging lens
PO =d*F/f₂ =15*120/30 =60 cm.
If the object is placed on the focus F₁ on the side of the diverging lens, then PF₁ =120 cm. The distance of the object from the diverging lens
OF₁ = PF₁-PO =120-60 =60 cm
70. A 5 mm high pin is placed at a distance of 15 cm from a convex lens of focal length 10 cm. A second lens of focal length 5 cm is placed 40 cm from the first lens and 55 cm from the pin. Find (a) the position of the final image, (b) its nature and (c) its size.
ANSWER: u = -15 cm, f =10 cm, v =?
1/v - 1/(-15) =1/10
→1/v =1/10-1/15 =5/150 =1/30
→v =30 cm
The image is formed on the other side of the convex lens. On this side, another lens at 40 cm from the first lens is kept. This image is now the object for the second lens. Object distance u =40 cm - 30 cm = 10 cm. Since the focal length of this lens is f = 5 cm, hence the object is placed at 2f =10 cm. For a convex lens, an object placed at 2f distance forms an image at the 2f distance on the other side of the lens.
(a) Hence the final image is at 2f = 10 cm from the second lens further away.
(b) Nature of the image:-
For the first lens, magnification m =v/u =hᵢ/hₒ
→hᵢ/hₒ =30/(-15) =-2
→hᵢ =-2hₒ =-2*5 mm =-10 mm
So the image is real and inverted.
For the second lens,
m =hᵢ/hₒ=v/u =2f/(-2f) =-1
→hᵢ = -hₒ =-10 mm
So the image is real and inverted and of the same size. Thus the first lens inverts the image and the second lens makes the final image erect by inverting again. So the final image is real and erect.
(c) Size of the image:-
As we saw in (b), the first lens makes the image size =10 cm and the second lens also makes the same size of the image, i.e. =10 cm.
ANSWER: First Method
Let the object be at x cm from the lens. Taking the origin at the optical center of the lens.
u = -x cm, f = -20 cm
1/v - 1/(-x) = 1/(-20)
→1/v = -1/20 - 1/x = -(x+20)/20x
→v =-20x/(x+20) -------------------(i)
So this virtual image is towards the object but it is the object for the mirror. Now taking the origin at the center of the mirror. Object distance
u = -5 -20x/(x+20) =-(5x+100+20x)/(x+20)
=-(25x+100)/(x+20)
f =-10 cm
From the mirror formula
1/v+1/u =1/f
→1/v-(x+20)/(25x+100) = -1/10
→1/v = (x+20)/(25x+100) - 1/10
→1/v =(10x+200-25x-100)/10(25x+100)
→1/v = (10-1.5x)/(25x+100)
→v = -(25x+100)/(1.5x-10)
 |
| Diagram for Q-61 |
This image distance is from mirror center C and in front of the mirror because the sign is negative. Again to form the image at the object the rays will go for second refraction through the lens and this image will be object for the lens. Assuming that the image is between the lens and the mirror (now origin will be taken at O again),
Now u =-[5-(25x+100)/(1.5x-10)]
{The value of v is taken without sign i.e. numerical value}
→u =-(7.5x-50-25x-100)/(1.5x-10)
=(17.5x+150)/(1.5x-10)
v = x
f =-20 cm
From the lens formula
1/v -1/u = 1/f
→1/x -(1.5x-10)/(17.5x+150) = -1/20
→(17.5x+150-1.5x²+10x)/x(17.5x+150) =-1/20
→20(-1.5x²+27.5x+150) =-17.5x²-150x
→-30x²+17.5x²+550x+150x+3000 =0
→-12.5x²+700x+3000=0
→x²-56x-240 =0
→x ={56士√(56²+4*240)}/2
={56士64}/2
=120/2 or -8/2
=60 or -4
In the given condition we neglect the negative value because it will be between the lens and the mirror and not real. So we have x = 60 cm i.e. the object should be placed 60 cm from the lens further away from the mirror.
Alternately
If the rays after reflection from the mirror trace back the same path, then the real image will be on the object itself. This will happen when the rays fall on the mirror perpendicularly. The rays will fall perpendicularly on the mirror if they will appear to come from the center of curvature.
 |
| Diagram for Q-61 |
Thus in the figure below, CR = radius of curvature =2f =2*10 =20 cm.
So, OR =20-5 =15 cm = image distance for the first refraction. But from (i) it is equal to 20x/(x+20). Equating these two,
20x/(x+20) =15
→20x =15x+300
→5x =300
→x =300/5 =60 cm.
62. A converging lens of focal length 12 cm and a diverging mirror of focal length 7.5 cm are placed at 5.0 cm apart with their principal axis coinciding. Where should an object be placed so that its image falls on itself?
ANSWER: For the lens f = 12 cm, assume the object is at distance x cm from the lens, so u = -x. From the lens formula
1/v-1/u =1/f
→1/v + 1/x =1/12
→1/v = 1/12 - 1/x =(x-12)/12x
→v = 12x/(x-12)
This image is the object for the mirror. If the rays after the first refraction fall perpendicularly on the mirror, then they will trace back and form the image on the object itself. This condition will be fulfilled if this image is formed on the center of curvature of the mirror R.
 |
| Diagram for Q-62 |
Thus OR = v =12x/(x-12)
→OC+CR =12x/(x-12)
{CR =radius of curvature =2f}
→5 + 2f =12x/(x-12)
→5+2*7.5 =12x/(x-12)
{for mirror, f =7.5 cm}
→12x/(x-12) =20
→12x =20x-240
→8x = 240
→x =240/8 =30 cm
So the object should be placed 30 cm from the lens further away from the mirror.
63. A converging lens and a diverging mirror are placed at a separation of 15 cm. The focal length of the lens is 25 cm and that of the mirror is 40 cm. Where should a point source be placed between the lens and the mirror so that the light, after getting reflected by the mirror and then getting transmitted by the lens, comes out parallel to the principal axis?
ANSWER: The rays come out parallel to the principal axis after refraction from a lens if the object is at its focus F. So in this case after reflection from the mirror the virtual image will be at F.
 |
| Diagram for Q-63 |
Suppose the object is placed at P at a distance x from the lens, OP =x. For the mirror, u =-(15-X), f =40 cm. From the mirror formula
1/v + 1/u = 1/f
→1/v-1/(15-x) =1/40
→1/v =1/(15-x) + 1/40 =(40+15-x)/{40*(15-x)}
→1/v =(55-x)/{40(15-x)}
→v = (600-40x)/(55-x) =CF =OF-OC =25-15 =10
{OF = focal length of the lens =25 cm}
→600-40x =550-10x
→30x =50
→x = 50/30 =1.67 cm
So the object should be kept at 1.67 cm from the lens.
64. A converging lens of focal length 15 cm and a converging mirror of focal length 10 cm are placed 50 cm apart with a common principal axis. A point source is placed in between the lens and the mirror at a distance of 40 cm from the lens. Find the locations of the two images formed.
ANSWER: One image will be formed after the refraction from the lens while the other will be formed first from the reflection of the mirror and then refraction from the lens.
From the given data the point source is on the focus of the converging mirror. After reflection from the mirror, the rays will go parallel and fall on the lens. These parallel rays will make an image at the focus of the lens i.e. at 15 cm from the lens on the other side.
For the image formed only by the lens, u = -40 cm, f= 15 cm, v =? From the lens formula
1/v - 1/u = 1/f
→1/v = 1/15 -1/40 =(8-3) /120 =5/120 =1/24
→v = 24 cm on the other side of the lens.
65. Consider the situation described in the previous problem. Where should a point source be placed on the principal axis so that the two images form at the same place?
ANSWER: Suppose such place is at a distance of x cm from the lens between them. After direct refraction from the lens, let the image be at y cm from the lens. In the second case where the first reflection from the mirror takes place if the image is formed at the object then after refraction from the lens again the final image will be at the place where direct refraction image was formed i.e. at y cm from the lens. Now the condition that the image formed by the mirror is at the object itself is that the object is at the radius of the curvature of the mirror. i.e. at 2f =2*10 cm =20 cm from the mirror. So the object should be placed at 50-20 =30cm from the lens towards the mirror.
66. A converging lens of focal length 15 cm and a converging mirror of focal length 10 cm are placed 50 cm apart. If a pin of length 2.0 cm is placed 30 cm from the lens farther away from the mirror, where will the final image form and what will be the size of the final image?
ANSWER: For the lens f=15 cm. Hence R=2f=30 cm.
The pin is placed at R of the converging lens. We know that for a converging lens if the object is placed at R, the inverted, real and same size image forms at R on the other side. In this case, the object AR will have an image R'D at 30 cm from the lens. It will be of the same size but inverted. This image will act as an object for the mirror. The focal length of the mirror f = 10 cm, Radius of curvature =CR' =2*10 = 20 cm. Since the distance between lens and mirror =OC= 50 cm, the object is 50-30 =20 cm from the mirror i.e. it is at the radius of curvature of the converging mirror. Its real image of the same size will be formed at the same place but inverted ER'. See the figure below:-
 |
| Diagram for Q-66 |
For the second refraction through the lens, this image is an object placed at the radius of curvature. So, again the same size, a real and inverted image on the other side of the lens at the 2f=30 cm distance will be formed i.e. at the real object itself.
67. A point object is placed on the principal axis of a convex lens (f = 15 cm) at a distance of 30 cm from it. A glass plate (µ = 1.50) of thickness 1 cm is placed on the other side of the lens perpendicular to the axis. Locate the image of the point object.
ANSWER: Since the object is placed at the 2f =2*15=30 cm distance from the lens, hence the image will be formed at 2f = 30 cm distance from the lens on the other side. If the 1 cm thick glass is placed within this distance there will be a shift in the image =(1-1/µ)*t =(1-1/1.5)*1 cm
=0.5/1.5 cm =0.33 cm.
Hence the image will be at 30+0.33 =30.33 cm from the lens.
68. A convex lens of focal length 20 cm and a concave lens of focal length 10 cm are placed 10 cm apart with their principal axes coinciding. A beam of light traveling parallel to the principal axis and having a beam diameter 5.0 mm, is incident on the combination. Show that the emergent beam is parallel to the incident one. Find the beam diameter of the emergent beam.
ANSWER: When the beam of light is incident from the side of the concave lens:-
 |
| Diagram for Q-68 |
Due to the position of the lenses, F is the common focus. PA is the topmost ray parallel to the principal axis. After refraction from the concave lens, it will appear to be coming from F and go along AB, i.e. the virtual image will be formed at the focus F. It can also be found out by the lens formula,
1/v +1/∞ =-1/10 {u =-∞, f =-10 cm}
→1/v =-1/10
→v = -10 cm =f
This virtual image will act as an object for the convex lens. Now F is also the focus of the convex lens. The rays coming from the focus of a convex lens go parallel to the principal axis. Hence the ray AB will go along BD after refraction from the convex lens, so the final image will be at infinity. It can also be found out from the lens formula,
1/v -1/(-20)=1/20 {u =-20 cm, f =20 cm}
→1/v =1/20-1/20 =0
→v =∞.
To get the beam diameter we compare triangles FAO' and FBO. Both are similar triangles, hence the corresponding sides will be proportionate.
BO/AO' =OF/OF'
→BO =AO'*OF/OF' =(5 mm)*(20 cm/10 cm)
→BO =5 mm*2 =10 mm =1 cm.
When the beam is incident from the side of the convex lens:-
 |
| Diagram for Q-68 |
The topmost ray PA which is parallel to the principal axis will form an image at the Focus F after refraction from the convex lens. This image will act as an object for the concave lens. An object at the focus makes the image at infinity, we can check it from the lens formula. u = O'F =10 cm, f =-10 cm, v =?
1/v - 1/10 =-1/10
→1/v =0
→v =∞, so the rays emerge parallel to the principal axis after refraction from the concave lens. For the beam diameter of the emergent beam, we find that the triangles AOF and BO'F are similar. Hence the corresponding sides will be proportionate.
BO'/AO =O'F/OF
→BO' =AO*O'F/OF =(5 mm)*(10 cm/20 cm)
→BO' =5 mm*(1/2) =2.5 mm.
69. A diverging lens of focal length 20 cm and a converging lens of focal length 30 cm are placed 15 cm apart with their principal axes coinciding. Where should an object be placed on the principal axis so that its image is formed at infinity?
ANSWER: For the diverging lens f₁ = -20 cm,
For the converging lens f₂ = 30 cm,
the distance between the lenses d =15 cm, hence the equivalent focal length of the combined lens, F is given as
1/F =1/f₁+1/f₂-d/f₁f₂
→1/F =-1/20+1/30-15/(-20)*30
→1/F =1/30-1/20+1/40
→1/F =(4-6+3)/120 =1/120
→F =120 cm
The positive sign shows that the equivalent lens is a converging lens. This focal length is the distance from the point P on the principal axis where the equivalent single lens is to be kept. The distance of the point P behind the converging lens
PO' =d*F/f₁ =15*120/(-20) =-90 cm. Negative sign shows that this distance is in the front of the converging lens.
 |
| Diagram for Q-69 |
If the object is kept at the focus of a converging lens, the image is formed at infinity on the other side. If the object is placed on the focus F₂ on the side of the convex lens, then PF₂ = 120 cm. The distance of the object from the converging lens
O'F₂ =O'P+PF₂ =90+120 =210 cm.
For the object placed on the other side, the distance of point P behind the diverging lens
PO =d*F/f₂ =15*120/30 =60 cm.
 |
| Diagram for Q-69 |
If the object is placed on the focus F₁ on the side of the diverging lens, then PF₁ =120 cm. The distance of the object from the diverging lens
OF₁ = PF₁-PO =120-60 =60 cm
70. A 5 mm high pin is placed at a distance of 15 cm from a convex lens of focal length 10 cm. A second lens of focal length 5 cm is placed 40 cm from the first lens and 55 cm from the pin. Find (a) the position of the final image, (b) its nature and (c) its size.
ANSWER: u = -15 cm, f =10 cm, v =?
1/v - 1/(-15) =1/10
→1/v =1/10-1/15 =5/150 =1/30
→v =30 cm
The image is formed on the other side of the convex lens. On this side, another lens at 40 cm from the first lens is kept. This image is now the object for the second lens. Object distance u =40 cm - 30 cm = 10 cm. Since the focal length of this lens is f = 5 cm, hence the object is placed at 2f =10 cm. For a convex lens, an object placed at 2f distance forms an image at the 2f distance on the other side of the lens.
(a) Hence the final image is at 2f = 10 cm from the second lens further away.
(b) Nature of the image:-
For the first lens, magnification m =v/u =hᵢ/hₒ
→hᵢ/hₒ =30/(-15) =-2
→hᵢ =-2hₒ =-2*5 mm =-10 mm
So the image is real and inverted.
For the second lens,
m =hᵢ/hₒ=v/u =2f/(-2f) =-1
→hᵢ = -hₒ =-10 mm
So the image is real and inverted and of the same size. Thus the first lens inverts the image and the second lens makes the final image erect by inverting again. So the final image is real and erect.
(c) Size of the image:-
As we saw in (b), the first lens makes the image size =10 cm and the second lens also makes the same size of the image, i.e. =10 cm.
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CHAPTER- 18 - Geometrical Optics
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CHAPTER- 15 - Wave Motion and Waves on a String
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CHAPTER- 14 - Fluid Mechanics
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CHAPTER- 2 - "Vector related Problems"
CHAPTER- 2 - "Vector related Problems"
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