In the following arrangements, O is a point source of light. The optical device is circular in shape and has aperture diameter D. D is very small so that all incident rays can be treated as paraxial. f is focal length of the device. A bright circular spot is formed on the screen in every case. Match the arrangements shown in column A with the diameter of spot given in column B.
List-I
List-II
P)
1) 5D
Q)
2) 3D
R)
3) D
S)
4) $\frac{3 D}{2}$

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P-4; Q-1; R-3; S-2

P-4; Q-2; R-3; S-1

P-3; Q-1; R-4; S-2

P-3; Q-2; R-4; S-1

answer is B.

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Detailed Solution

(P) Rays will become parallel to principal axis after reflection. Therefore, diameter of spot will be D.
(Q) $\frac{1}{v} + \frac{1}{(- f)} = \frac{1}{(+ f)} \Rightarrow v = + \frac{f}{2}$
If diameter of the spot is y then $\frac{y}{(\frac{5 f}{2})} = \frac{D}{(\frac{f}{2})}$ or, y = 5D

(r) $\frac{1}{v} - \frac{1}{(- \frac{f}{2})} = \frac{1}{(+ f)} \Rightarrow v = - f$
Diameter of the spot is given by $\frac{y}{(\frac{3 f}{2})} = \frac{D}{f}$ or, $y = \frac{3 D}{2}$
Question Image
(S) Diameter of the spot is given by $\frac{y}{3 d} = \frac{D}{d}$ or $y = 3 D$