Wave Optics

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Q1.

In Young’s double-slit experiment, how many minimas can be obtained on a screen, if the wavelength of light used is 2000 $\overset{o}{A}$ and slit width is 0.0008 mm?

Explanation:

Path difference=dsin $θ = (2 n - 1) \frac{λ}{2}$

sin $θ$ $= (2 n - 1) \frac{λ}{2 d} = (2 n - 1) \frac{2000 \times 10^{- 10}}{2 \times 0.0008 \times 10^{- 3}} = \frac{(2 n - 1)}{8}$

$Maximum value of n = 4$
$So the total no . of minimas = 8 (4 above and 4 below the central maxima)$

Q2.

A beam of light of 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between first dark fringes on either side of the central bright fringe is

Explanation:

$F r o m t h e f i g u r e, \tan θ = \frac{\frac{x}{2}}{2} = \frac{x}{4}$
$F o r s m a l l θ a n d w h e n θ i s c o u n t e d i n r a d, \tan θ \approx θ \Rightarrow θ = \frac{x}{4}$
$\Rightarrow \frac{λ}{a} = \frac{x}{4}$
$\Rightarrow x = \frac{4 λ}{a} = 2.4 m m$