JEE Main 2016 · 9 Apr · Shift 1 · Physics Q20

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PhysicsSingle correct+4 / -1Question 20 of 90

In Young's double slit experiment, the distance between slits and the screen is 1.0 m and monochromatic light of 600 nm is being used. A person standing near the slits is looking at the fringe pattern. When the separation between the slits is varied, the interference pattern disappears for a particular distance $d_0$ between the slits. If the angular resolution of the eye is $\dfrac{1^\circ}{60}$ , the value of $d_0$ is close to :

1 mm

2 mm

4 mm

3 mm

Answer and worked solution

Official answer: B

The fringes can no longer be resolved when their angular separation $\frac\lambda d$ falls to $1'=\frac{\pi}{180\times60}=2.9\times10^{-4}$ rad. So $d_0=\frac{6\times10^{-7}}{2.9\times10^{-4}}\approx2$ mm.

Topics:Interference and Young's double slit Resolving power

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JEE Main 2016 · 9 Apr · Shift 1, Physics question 20