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Question

A parallel beam of light of wavelength 600 nm is incident normally on a slit of width 'a'. If the distance between the slit and the screen is 0.8 m and the distance of 2nd order maximum from the centre of the screen is 1.5  mm, calculate the width of the slit.


Solution

Given $$\lambda = 600 nm = 600 \times 10^{-9} m = 6.0 \times 10^{-7} m, D = 0.8 m$$,

$$y_2 = 1.5 mm = 1.5 \times 10^{-3} m, n = 2, a = ?$$

Position of $$n^{th}$$ maximum in diffraction of a single slit

$$y_n = (n + \dfrac{1}{2}) \frac{\lambda D}{a}  \Rightarrow  \ \ a = (n + \dfrac{1}{2}) \dfrac{\lambda D}{y_n}$$

Substituting given values $$a = (2 + \dfrac{1}{2}) \dfrac{6.0 \times 10^{-7} \times 0.8}{1.5 \times 10^{-3}}$$

$$a= \dfrac{5}{2} \times 4.0 \times 0.8 \times 10^{-4}m = 0.8 \times 10^{-3} m = 0.8 mm$$

Physics
NCERT
Standard XII

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