Question

A parallel beam of light falls successively on a thin convex lens of focal length 40 cm and then ono a thini convex lens of focal length 10cm as shown in figure.
In figure, the second lens is an equiconcave lens of focal length 10cm and made of a material of refractive index 1.5. In both the cases, the second lens has an aperture equal to 1cm.

Q. Compare the area illuminated by the beam of light on the screen, which passes through the second lens in the two cases. The ratio `(A_(2)//A_(1))` will be
A. `72//5`
B. `81//4`
C. `56//3`
D. `29//2`

Answer 1

Correct Answer - b.
In case (a), the incident parallel beam emerges as a parallel beam. So area illuminated,
`A_(1)=pi(1)^(2)=picm^(2)`
In case (b), let x be the diameter of the area illuminated.
Then,
`(x)/(45)=(1)/(5)rArr x=9cm`
`A_(2)=pi((9)/(2))^(2)=(81)/(4)picm^(2)`
`(A_(2))/(A_(1))=(81)/(4)`
When liquid of refractive index `mu` is filled to the right of this lens, the first surface of the lens (radius of curvature `=10cm)` forms the imag at the object only. Considering the refraction at the second surface.
`(mu)/(oo)-(1.5)/(-10)=(mu-1.5)/(10)` (therefore, same area `rArr upsilonrarroo)`
`rArr mu=3`