Monday, June 24, 2019

H C Verma solutions, Optical Instruments, EXERCISES, Q1_12, Chapter-19, Concepts of Physics, Part-I

OPTICAL INSTRUMENTS

EXERCISES -Q1_12


1. A person looks at different trees in an open space with the following details. Arrange the trees in decreasing order of their apparent sizes.

Tree         Height (m)      Distance from the eye(m)

A                 2.0                         50
B                 2.5                         80
C                 1.8                         70
D                 2.8                        100



Answer: The apparent size of a tree will depend on the angle subtended by the tree on the eye.

The angle subtended by A = 2.0/50 = 0.04
by B = 2.5/80 = 0.03
by C = 1.8/70 = 0.025
by D = 2.8/100 = 0.028
Larger the angle, larger the apparent size of the tree.
Hence they can be arranged as, A, B, D, C.

2. An object is to be seen through a simple microscope of focal length 12 cm. Where should the object be placed so as to produce maximum angular magnification? The least distance for clear vision is 25 cm.



Answer: The focal length of the lens, f = 12 cm,

For maximum angular magnification, image distance, v= -25 cm,
u =? 
From the lens formula,
1/v - 1/u = 1/f
→-1/25 - 1/u =1/12
→1/u =-1/25 - 1/12 = -(12+25)/300 =-37/300
→u = -300/37 = -8.1 cm
Negative sign shows that the object should be in front of the lens. The distance of the object from the lens = 8.1 cm.






3. A simple microscope has a magnifying power of 3.0 when the image is formed at the near point (25 cm) of a normal eye. (a) What is its focal length? (b) What will be its magnifying power if the image is formed at infinity?



Answer: (a) Given, m = 3.0, D =25 cm. Hence,

m = 1 + D/f
→3 = 1 + 25/f
→25/f = 2
→f = 25/2 = 12.5 cm.

(b) When the image is formed at infinity, the magnifying power is given as
m = D/f = 25/12.5 = 2.


  

4. A child has near point at 10 cm. What is the maximum angular magnification the child can have with a convex lens of focal length 10 cm?



Answer: The maximum angular magnification is obtained when the image is at the near point.

m = 1 +D/f
Here, D = 10 cm, f =10 cm.
m = 1 + 10/10 = 1+1 = 2.

5. A simple microscope is rated 5X for a normal relaxed eye. What will be its magnifying power for a relaxed farsighted eye whose near point is 40 cm?



Answer: Let the focal length of the lens = f.

For a normal relaxed eye, the image is formed at infinity. Hence the angular magnification is given as,
m = D/f.
Here, m = 5, D = 25 cm
So, 5 =25/f
→f = 25/5 = 5 cm.
For a relaxed farsighted person whose D = 40 cm,
The magnifying power of the microscope is
m = D/f = 40/5 = 8 X. 





6. Find the maximum magnifying power of a compound microscope having a 25 diopter lens as the objective, a 5 diopter lens as the eyepiece and the separation 30 cm between the two lenses. The least distance for clear vision is 25 cm.



Answer: The maximum magnifying power of a compound microscope is given when the final image is at D.

m = -(vₒ/uₒ){1 + D/fₑ}

Given, L = 30 cm, fₒ =1/25 m =4 cm, fₑ =1/5 m =20 cm, D = 25 cm.
For the eyepiece, v = -25 cm (virtual), fₑ = 20 cm, so
1/v - 1/ u = 1/f
→ -1/25 -1/u = 1/20
→1/u = -(1/25 + 1/20) = -45/500
→u = -500/45 = -11.11 cm
So the object is 11.11 cm in front of the eyepiece. It is infact the image formed by the objective. Distance of the image from the objective = L-11.11 
=30 -11.11 = 18.89 cm. It is a real image behind the objective, hence v = 18.89 cm, fₒ = 4 cm, from the lens formula
1/v - 1/u = 1/f
→1/18.89 - 1/u = 1/4
→1/u =1/18.89 - 1/4 =-0.197
→u = -5.08 cm
Diagram for Q-6

Now vₒ =18.89 cm, uₒ =-5.08 cm, So,
m = -{18.89/(-5.08)}{1 + 25/20}
    = 3.72*45/20
    = 8.37
    ≈ 8.4






7. The separation between the objective and the eyepiece of a compound microscope can be adjusted between 9.8 cm to 11.8 cm. If the focal lengths of the objective and the eyepiece are 1.0 cm and 6 cm respectively, find the range of the magnifying power if the image is always needed at 24 cm from the eye.




Answer: For the eyepiece, fₑ = 6 cm, for maximum magnification image should be at D, so vₑ = -24 cm, hence from the lens formula

-1/24 - 1/uₑ = 1/6
→1/uₑ = -1/24 - 1/6 = -30/144
→uₑ = -144/30 = -4.8 cm
So the object and hence the image of the objective is 4.8 cm in front of the eyepiece. Hence for the eyepiece
vₒ = L-4.8 =9.8-4.8 =5.0 cm
{If L = 9.8}
fₒ = 1.0 cm
uₒ =?
1/5 - 1/uₒ = 1/1
→1/uₒ = 1/5-1 =-4/5
→uₒ = -5/4 = -1.25
So the angular magnification, 
m =-(vₒ/uₒ){1 + D/fₑ}
→m =(5/1.25){1+24/6} =-4*30/6 =-120/6
→m = 20

If L is adjusted to 11.8 cm, then
vₒ = 11-4.8 =7.0 cm
fₒ = 1.0 cm, hence
1/7 - 1/uₒ =1/1 = 1
→1/uₒ =1/7-1 =-6/7
→uₒ = -7/6 cm
Now the angulr magnification
m =(7*6/7){1 + 24/6} =6*30/6 =30

Hence the range of the magnifying power will be between 20 to 30.


8. An eye can distinguish between two points of an object if they are separated by more than 0.22 mm when the object is placed at 25 cm from the eye. The object is now seen by a compound microscope having a 20 D objective and 10 D eyepiece separated by a distance of 20 cm. The final image is formed at 25 cm from the eye. What is the maximum separation between two points of the object which can now be distinguished?




Answer: We actually need to know the angular magnification of the compound microscope.

fₑ = 1/10 m =10 cm, vₑ =-25 cm, uₑ =? From the lens formula,
-1/25 - 1/uₑ = 1/10
→1/uₑ =-1/25 - 1/10 =-35/250
→uₑ = -250/35 =-7.14 cm
So the object is at 7.14 cm in front of the eyepiece.
For the objective, vₒ = L-uₑ =20 - 7.14 =12.86 cm
fₒ = 1/20 m =5 cm, uₒ = ?
From the lens formula,
1/12.86 -1/uₒ = 1/5
→1/uₒ = 1/12.86 -1/5 = -0.12
→uₒ = -8.33

Angular magnification, m =-(vₒ/uₒ){1+D/fₑ}
→m =(12.86/8.33){1 +25/10} =1.54*3.5
→m =5.39

The minimum distance that can be distinguished without microscope = 0.22 mm
Hence the minimum distance that can be distinguished with this microscope = 0.22/5.39 
=0.04 mm





9. A compound microscope has a magnifying power of 100 when the image is formed at infinity. The objective has a focal length of 0.5 cm and the tube length is 6.5 cm. Find the focal length of the eyepiece.




Answer: Given m =100 when the image is at infinity.

In this case, the image by the objective is at the focus of the eyepiece. L = 6.5 cm, vₒ =6.5-fₑ 
fₒ = 0.5 cm, uₒ =? from the lens formula
1/vₒ - 1/uₒ = 1/0.5
→1 - vₒ/uₒ = 2vₒ
{multiplying both sides by vₒ}
→vₒ/uₒ = 1 - 2vₒ

Also the magnification, m = -(vₒ/uₒ)(D/fₑ)
→100 = -{1-2(6.5-fₑ)}{25/fₑ}
→100/25 = -{2fₑ - 12}/fₑ
→4 = -2{1- 6/fₑ}
→-2 = 1 - 6/fₑ
→6/fₑ = 3
→fₑ = 6/3 = 2.






10. A compound microscope consists of an objective of focal length 1 cm and an eyepiece of focal length 5 cm. An object is placed at a distance of 0.5 cm from the objective. What should be the separation between the lenses so that the microscope projects an inverted real image of the object on a screen 30 cm behind the eyepiece?




Answer: Given, fₑ = 5 cm, fₒ = 1 cm, uₒ = -0.5 cm, vₑ = 30 cm. For the objective from the lens formula,

1/vₒ - 1/(-0.5) = 1/1
→1/vₒ = 1-2 =-1
→vₒ = -1 cm
So the image is 1 cm in front of the objective. This image will act as the object for the eyepiece. Let the separation between the lenses of the microscope = L.

Now, uₑ = -(L+1) cm, fₑ = 5 cm, vₑ = 30 cm (given). From the lens formula,
1/30 - 1/{-(L+1)} =1/5
→1/(L+1) = 1/5 -1/30 =(6-1)/30 =1/6
→L+1 = 6
→L = 6-1 =5 cm.





11. An optical instrument used for angular magnification has 25 D objective and a 20 D eyepiece. The tube length is 25 cm when the eye is least strained.

(a) Whether it is a microscope or a telescope?
(b) What is the angular magnification produced?



Answer: From the given data, 

fₒ = 1/25 D =100/25 cm =4 cm.

fₑ = 1/20 D =100/20 cm =5 cm

(a) Since fₒ < fₑ, hence the instrument is a microscope.

(b) When the eye is least strained, the image by the eyepiece is at infinity and the object (image by the objective) is at the focus of the eyepiece. Given, L = 25 cm, hence vₒ = L - fₑ =25 - 5 =20 cm. uₒ =?
From the lens formula,
1/vₒ - 1/uₒ = 1/fₒ
→1/20 - 1/uₒ = 1/4
→1/uₒ = 1/20 - 1/4 =(1-5)/20 = -1/5
→uₒ = -5 cm
Hence the angular magnification produced,
m = -(vₒ/uₒ)(D/fₑ) =(20/5)(25/5) =4*5 =20.






12. An astronomical telescope is to be designed to have a magnifying power of 50 in normal adjustment. If the length of the tube is 102 cm, find the powers of the objective and the eyepiece.




Answer: Given m = 50. In normal adjustment, the image by the eyepiece is at infinity. It means the object (image produced by the objective) is at the focus of the eyepiece. Since the object for the objective is at infinity, the image is at its focus, i.e. vₒ=fₒ. Given L = 102 cm,

Hence fₒ + fₑ = 102 ----------(i)

In this case the angular magnification, m = fₒ/fₑ.
→50 = fₒ/fₑ
→fₒ = 50fₑ
Putting it in (i)
50fₑ +fₑ = 102
→51fₑ = 102
→fₑ = 2 cm = 0.02 m
And fₒ = 50*2 =100 cm = 1 m

Hence the power of the objective =1/fₒ =1/1 = 1 D
→the power of the eyepiece = 1/fₑ =1/0.02 =50 D.




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