Physics, asked by satendraseemasharawa, 5 months ago

22) A luminous object & a screen are placed on an optical bench & a converging lens is placed
between them to through a sharp image of the object on the screen the linear
magnification of the image is found to be 2.5. The lens is now moved 30cm nearer the
screen & a sharp image is again form. Calculate the
focal length of length of the lens.
DIC
- 20 cm​

Answers

Answered by rahulpandey4454
0

Answer:

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12th

Physics

Ray Optics and Optical Instruments

Image Formation by Lenses

An object and a screen are ...

PHYSICS

An object and a screen are mounted on an optical bench and a converging lens is placed between them so that a sharp image is received on the screen. The linear magnification of the image is 2.5. The lens is now moved 30 cm nearer to the screen and a sharp image is again formed on the screen. The focal length of the lens is:

A

14.0cm

B

14.3cm

C

14.6cm

D

14.9cm

HARD

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VIDEO EXPLANATION

12th

Physics

Ray Optics and Optical Instruments

Image Formation by Lenses

An object and a screen are ...

PHYSICS

An object and a screen are mounted on an optical bench and a converging lens is placed between them so that a sharp image is received on the screen. The linear magnification of the image is 2.5. The lens is now moved 30 cm nearer to the screen and a sharp image is again formed on the screen. The focal length of the lens is:

A

14.0cm

B

14.3cm

C

14.6cm

D

14.9cm

HARD

Share

Study later

VIDEO EXPLANATION

ANSWER

Given uv=2.5

v=2.5u

again v−u=30

v=30+u

2.5u=30+u

1.5u=30

u=20

Now, f1=u1+v1=uvu+v

f=u+vuv=3.5u2.5u2

   =75u=75×20=7100=14.3cm

Answered by awakashkumarray2007
0

Answer:

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