JEE Main · 2016 · 03 Apr 2016 · Offline · Wave Optics

The box of pin hole camera, of length L, has a hole of radius

a

. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength

λ

the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say

b_{min}

) when :

(A) $a = λ L$ and $b_{min} = (\frac{2 λ ^{2}}{L})$
(B) $a = λ L$ and $b_{min} = 4 λ L$ ✓ Correct
(C) $a = \frac{λ ^{2}}{L}$ and $b_{min} = 4 λ L$
(D) $a = \frac{λ ^{2}}{L}$ and $b_{min} = (\frac{2 λ ^{2}}{L})$

Answer: (B)

Solution

The spot is widened by two competing effects: the geometric shadow of the hole, of size about

a

, and diffraction, of size about

\frac{λ L}{a}

. Adding them,

b = a + \frac{λ L}{a}

. To make the spot as small as possible, minimise this:

\frac{d b}{d a} = 1 - \frac{λ L}{a ^{2}} = 0

gives

a = λ L

. Putting this back,

b_{m i n} = λ L + λ L = 2 λ L

, which is the same as

4 λ L

JEE Main · 2016 · 03 Apr 2016 · Offline · Single correct option

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JEE Main · 2016 · 03 Apr 2016 · Offline · Wave Optics

Solution

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