Question

A point source of light $S$ is placed on the axis of a lens of focal length $20cm$ as shown . A screen is placed normal to the axis of lens at a distance $x$ from it . Treat all rays as paraxial :

• A. As $x$ is increases from zero intensity continuously decreases
• B. As $x$ is increased from zero intensity first increases then decreases
• C. Intensity at centre of screen for $x=90cm$ and $x=110cm$ is same
• D. Radius of bright circle obtained on screen is equal to $1cm$ for $x=200cm$

Answer 1

Answer: (B), (C), (D)

As $x$ is increased from zero intensity first increases then decreases
Intensity at centre of screen for $x=90cm$ and $x=110cm$ is same
Radius of bright circle obtained on screen is equal to $1cm$ for $x=200cm$

we know lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

in given question $u=-25cm,f=20cm$.

substituting value in lens formula we get

$\frac{1}{v}-\frac{1}{-25}=\frac{1}{20}$

$\frac{1}{v}=\frac{1}{20}-\frac{1}{25}$

$\frac{1}{v}=\frac{5-4}{100}$

$\frac{1}{v}=\frac{1}{100}$

$v=100cm$ and + sign signifies at right side of lens.

brightest image formed on screen when $x=100cm$

when $x=0$, we can't see any image and when $x=\infty$, we can't see any image.

so intensity at $x=0$ is low then intensity increases till $x=100cm$,

here intensity is maximum then again intensity decreases.

so intensity at equal distance from $x=100cm$ in both direction is same.

from image you can say that height of image formed is $2cm$.

Diameter of ring is $2cm$ i.e radius of ring is $1cm$.

Option(A) is wrong Option (B) ,(C) and (D) are correct.