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Physics Question on Wave optics

In a double slit experiment using light of wavelength 600 nm600\, nm, the angular width of a fringe on a distant screen is 0.1∘0.1^{\circ}. The spacing between the two slits is

A

3.44×10−4 m3.44 \times 10^{-4}\,m

B

1.54×10−4 m1.54 \times 10^{-4}\,m

C

1.54×10−3 m1.54 \times 10^{-3}\,m

D

1.44×10−3 m1.44 \times 10^{-3}\,m

Answer

3.44×10−4 m3.44 \times 10^{-4}\,m

Explanation

Solution

Angular width of fringe, θ=0.1∘=0.1×π180rad\theta = 0.1^{\circ} = 0.1 \times\frac{\pi}{180} rad =3.141800rad= \frac{3.14}{1800}rad ∴d=λθ\therefore d = \frac{\lambda}{\theta} =600×10−93.14/1800 = \frac{600\times 10^{-9}}{3.14/ 1800} =3.44×10−4m= 3.44 \times 10^{-4} m