Question

A double slit is illuminated by light of wavelength $6000\stackrel{\circ }{A}$. The slits are $0.1cm$ apart and the screen is placed one meter away. Then the angular position of the ${10}^{th}$ maximum in radian and separation of the two adjacent minima will be respectively

• A. $6\times {10}^{-3}\text{rad},0.6mm$
• B. $3\times {10}^{-3}\text{rad},0.3mm$
• C. $2\times {10}^{-3}\text{rad},0.2mm$
• D. $4\times {10}^{-3}\text{rad},0.4mm$

Answer 1

The correct option is A $6\times {10}^{-3}\text{rad},0.6mm$


Distance of $n^{th}$ maximum from centre
$y=\frac{n\lambda D}{d}$

So, angular position of $n^{th}$ maximum
$\theta =\frac{y}{D}=\frac{n\lambda }{d}$ (small angle approximation)

So for $n=10$, ${\theta }_n=\frac{10\times 6000\times {10}^{-10}}{0.001}=6\times {10}^{-3}\text{rad}$

Separation between two adjacent minima = separation between two adjacent maxima = fringe width
$\beta =\frac{\lambda D}{d}=\frac{6000\times {10}^{-10}\times 1}{0.001}=6\times {10}^{-4}m=0.6mm$