Saturday, June 29, 2019

H C Verma solutions, Optical Instruments, EXERCISES, Q13_24, Chapter-19, Concepts of Physics, Part-I

OPTICAL INSTRUMENTS

EXERCISES -Q13_24


13. The eyepiece of an astronomical telescope has a focal length of 10 cm. The telescope is focused for the normal vision of distant objects when the tube length is 1.0 m. Find the focal length of the objective and the magnifying power of the telescope.



Answer: The fₑ = 10 cm. L = 1.0 m =100 cm, fₒ =? 

For the normal vision, the final image by the eyepiece is at infinity. Since the object is far away from the objective, the image by the objective is at its focus which is also the focus of the eyepiece. Hence the length of the telescope,
L = fₒ + fₑ
→100 =fₒ + 10
→fₒ = 90 cm.
Diagram for Q-13

The magnifying power is given as,
m = fₒ/fₑ = 90/10 = 9. 


14. A Galilean telescope is 27 cm long when focused to form an image at infinity. If the objective has a focal length of 30 cm, what is the focal length of the eyepiece?




Answer: The focal length of the objective, fₒ = 30 cm. Length of the Galilean telescope, L = 27 cm. fₑ =?

In a Galilean telescope, the converging objective forms the image at its focus because the object is at far away. The eyepiece, which is a diverging lens, is adjusted such that this image is at its focus so that the final virtual image is formed at infinity. Thus the focal length of the objective is greater than the length of the telescope since its focus is behind the eyepiece. Hence the length, L = fₒ - fₑ
→27 = 30 - fₑ
→fₑ = 30 - 27 = 3 cm.







15. A farsighted person cannot see objects placed closer to 50 cm. Find the power of the lens needed to see the objects at 20 cm.




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

Object distance, u = -20 cm, Image distance v = - 50 cm. Hence from the lens formula,
1/v - 1/u = 1/f
→1/f = 1/20 - 1/50 =(5-2)/100 = 3/100
→f = 100/3 cm =1/3 m.
Hence the power of the lens P = 1/f =3 D.




16. A nearsighted person cannot clearly see beyond 200 cm. Find the power of the lens needed to see objects at large distances.




Answer: The person needs a lens such that the image of objects is formed at 200 cm. Hence, u = -∞, v =-200 cm, f = ? From the lens formula,

1/v - 1/u = 1/f
→1/f = 1/∞ - 1/200 =-1/200
→f = -200 cm = -2 m

Hence the power of the lens, 
P = 1/f = -1/2 D = -0.5 D.



17. A person wears glasses of power - 2.5 D. Is the person farsighted or nearsighted? What is the far point of the person without the glasses?




Answer: Since the power of the lens is negative, it is a diverging lens and the person is nearsighted.

Let the far point of the person = x. It means the image of objects far away is formed at x distance from the lens. So, u = -∞, f = -1/2.5 m =100/2.5 cm =1000/25 cm = 40 cm. From the lens formula,
1/v - 1/u = 1/f
→1/x + 1/∞ =1/f
→1/x = 1/f
→ x = f = 40 cm.




18. A professor reads a greeting card received on his 50th birthday with 2.5 D glasses keeping the card 25 cm away. Ten years later, he reads his farewell letter with the same glasses but he has to keep the letter 50 cm away. What power of lens should he now use?




Answer: Let the image of the greeting card be at x distance. u = -25 cm, f = 1/2.5 m =100/2.5 cm =40 cm. v = -x. From the lens formula,

1/v - 1/u = 1/f
→1/25 - 1/x = 1/40
→1/x = 1/25 - 1/40 =(8 - 5)/200 =3/200
→x = 200/3 cm
{It is his near point}
For the farewell letter, f = 40 cm, u = -50 cm, v = ?
so, 
1/v + 1/50 = 1/40
→1/v = 1/40 - 1/50 =(5 - 4)/200 =1/200
→v = 200 cm
So his near point is 200 cm.

The requirement of the lens:-
The lens should make the image at 200 cm if the object i.e. farewell letter is placed at 25 cm. Let the focal length of the lens = f. So, u = -25 cm, v = 200 cm. Now
1/f = 1/v - 1/ u =1/200 + 1/25 = 9/200
→f = 200/9 cm = 2/9 m
Hence the power of the lens required =1/f =9/2 =4.5 D.





19. A normal eye has retina 2 cm behind the eye-lens. What is the power of the eye lens when the eye is (a) fully relaxed, (b) most strained?


Answer: (a) When the normal eye is relaxed, it is focused on objects situated at a large distance. So, u = -∞. The image is formed at the retina, so, v = 2 cm. Let focal length = f. From the lens formula,

1/v -1/u = 1/f
→1/2 +0 = 1/f
→f = 2 cm =0.02 m
Hence the power of the eye-lens, 
P = 1/f =1/0.02 D =50 D.

(b) When the normal ye is most strained, it is seeing the object at the least distance of clear vision which is 25 cm. So, u = -25 cm, v = 2 cm. From the lens formula,
1/2 +1/25 =1/f
→1/f =(25+2)/50 =27/50
→f =50/27 cm =1/54 m
Hence the power of the eye lens,
P =1/f =54 D.



20. The near point and the far point of a child are 10 cm and 100 cm. If the retina is 2.0 cm behind the eye-lens, what is the range of the power of the eye-lens?



Answer: Since the image is formed at the retina, the image distance will be the same, i.e. v = 2 cm. When the object is at the near point, u = -10 cm. Hence from the lens formula
1/f = 1/2 + 1/10 =6/10 
→f =10/6 cm =1/60 m
Hence the power of the eye lens,
P =1/f =60 m.
When the object is at the far point, u =-100 cm.
From the lens formula,
1/f = 1/2 + 1/100 
→1/f = 51/100
→f =100/51 cm =1/51 m
Hence the power of the eye lens,
P =1/f =51 D.
So the range of the power of the eye lens is from +60 D to +51 D.


21. A nearsighted person cannot see beyond 25 cm. Assuming that the separation of the glass from the eye is 1 cm, find the power of the lens needed to see distant objects.


Answer: For the correction of the near sighting problem the image of the objects at infinity should be formed at 25 cm from the eye. Since the separetion of the glass from the eye is1 cm, this image distance from the lens, v = -(25-1) =-24 cm. u =-∞. Hence from the lens formula,

1/f =1/v - 1/u = -1/24 +0 =-1/24
→f = -24 cm = -0.24 m
hence the power of the lens,
P = -1/0.24 = -4.2 D.



22. A person has near point at 100 cm. What power of lens is needed to read at 20 cm if he/she uses (a) contact lens, (b) spectacles having glasses 2.0 cm separated from the eyes?


Answer: (a) When the contact lens is used, there is no separation between the eye and the lens. Hence,

 Image distance, v = -100 cm = -1 m, 

Object distance, u = -20 cm = -0.20 m, {Given}. Hence from the lens formula,

1/f = -1/1 + 1/0.20 = -1 + 5 =4
Hence the power of the lens required,
P = 1/f = 4 D.

(b) If the person uses spectacles, having separation from the eye = 2.0 cm, the image and the object distances from the glass change. Now, u = -(20-2) =-18 cm =-0.18 m, v = -(100-2) =-98 cm =0.98 m. Hence from the lens formula,

1/f = -1/0.98 + 1/0.18 =-1.02 + 5.55 = 4.53

Hence the person  needs glasses of power, 
P = 1/f = 4.53 D. 



23. A lady uses +1.5 D glasses to have normal vision from 25 cm onwards. She uses a 20 D lens as a simple microscope to see an object. Find the maximum magnifying power if she uses the microscope (a) together with her glass (b) without the glass. Do the answers suggest that an object can be more clearly seen through a microscope without using the correcting glasses?


Answer: (a) Microscope's focal length, f = 1/P =1/20 m =100/20 cm =5 cm. Together with glasses, the person's near vision is normal. Hence


 D =25 cm.

Hence the magnifying power of the lens
m = 1 +D/f = 1 + 25/5 = 6

(b) To know the magnification without the glass we need to know the D (minimum distance of the clear vision) for the eye lens.
Power of the glass used P = +1.5 D
f = 1/P =1/1.5 m = 2/3 m =200/3 cm.
This glass forms the image at D when the object is at 25 cm. Hence, u = -25 cm, v = -D. From the lens formula,
1/v - 1/u = 1/f
→-1/D + 1/25 =3/200
→1/D = 1/25 - 3/200 =5/200 =1/40
→D = 40 cm.
Now the maximum magnification of the simple microscope = m = 1 +D/f =1 + 40/5 =1+8 = 9.


24. A lady cannot see objects closer than 40 cm from the left eye and closer than 100 cm from the right eye. While on a mountaineering trip, she is lost from her team. She tries to make an astronomical telescope from her reading glasses to look for her teammates. (a) Which glass should she use as the eyepiece? (b) What magnification can she get with a relaxed eye?



Answer: The left lens makes an image at 40 cm for the object at 25 cm. Hence u = -25 cm, v = -40 cm. Hence

1/f = -1/40 + 1/25 = (-5+8)/200 =3/200

→f = 200/3 cm =66.66 cm

Similarly the right lens makes an image at 100 cm fo the object placed at 25 cm. Hence u = -25 cm, v = -100 cm. From the lens formula,

1/f = -1/100 + 1/25 = (-1 + 4)/100 =3/100

f = 100/3 cm = 33.33 cm

(a) In an astronomical telescope, the objective is kept of larger focal length while the eyepiece has the smaller focal length. Since the right lens has the smaller focal length, it should be used as the eyepiece. 

(b) Now fₒ = 66.66 cm, fₑ = 33.33 cm. With the relaxed eyes the final image is at infinity and it is the normal adjustment. In this case the magnification

m = fₒ/fₑ =66.66/33.33 =2.

 ===<<<O>>>===



click here for all links → kktutor.blogspot.com 








===<<<O>>>===



My Channel on YouTube  →  SimplePhysics with KK

Links to the Chapters




CHAPTER- 14 - Fluid Mechanics



CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion








CHAPTER- 11 - Gravitation




CHAPTER- 10 - Rotational Mechanics




CHAPTER- 9 - Center of Mass, Linear Momentum, Collision


CHAPTER- 8 - Work and Energy

Click here for → Question for Short Answers

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

CHAPTER- 7 - Circular Motion

Click here for → Questions for Short Answer 

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → EXERCISES (11-20)

Click here for → EXERCISES (21-30)

CHAPTER- 6 - Friction

Click here for → Questions for Short Answer

Click here for → OBJECTIVE-I

Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

For more practice on problems on friction solve these- "New Questions on Friction".

---------------------------------------------------------------------------------

CHAPTER- 5 - Newton's Laws of Motion


Click here for → QUESTIONS FOR SHORT ANSWER

Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

-------------------------------------------------------------------------------

CHAPTER- 4 - The Forces

The Forces-

"Questions for short Answers"    


Click here for "The Forces" - OBJECTIVE-I


Click here for "The Forces" - OBJECTIVE-II


Click here for "The Forces" - Exercises


--------------------------------------------------------------------------------------------------------------

CHAPTER- 3 - Kinematics - Rest and Motion

Click here for "Questions for short Answers"


Click here for "OBJECTIVE-I"


Click here for EXERCISES (Question number 1 to 10)


Click here for EXERCISES (Question number 11 to 20)


Click here for EXERCISES (Question number 21 to 30)


Click here for EXERCISES (Question number 31 to 40)


Click here for EXERCISES (Question number 41 to 52)


CHAPTER- 2 - "Vector related Problems"

Click here for "Questions for Short Answers"


Click here for "OBJECTIVE-II"



Posted by at No comments:
Labels: , ,
Subscribe to: Posts (Atom)