JEE Main 2016 · 3 Apr · Shift 1 · Physics Q20

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PhysicsSingle correct+4 / -1Question 20 of 90

The box of a 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=\dfrac{\lambda^2}L

and

b_{min}=\left(\dfrac{2\lambda^2}L\right)

a=\sqrt{\lambda L}

and

b_{min}=\left(\dfrac{2\lambda^2}L\right)

a=\sqrt{\lambda L}

and

b_{min}=\sqrt{4\lambda L}

a=\dfrac{\lambda^2}L

and

b_{min}=\sqrt{4\lambda L}

Answer and worked solution

Official answer: C

The geometric spread is about the hole size $a$ , and diffraction adds about $\dfrac{\lambda L}a$ , so $b\approx a+\dfrac{\lambda L}a$ . This is smallest when $a=\sqrt{\lambda L}$ , giving $b_{min}=2\sqrt{\lambda L}=\sqrt{4\lambda L}$ .

Topics:Diffraction

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JEE Main 2016 · 3 Apr · Shift 1, Physics question 20