wave-optics Question 22
Question: Q. 5. (i) In Young’s double slit experiment, two slits are $1 \mathrm{~mm}$ apart and the screen is placed $1 \mathrm{~m}$ away from the slits. Calculate the fringe width when light of wave length $500 \mathrm{~nm}$ is used ? (ii) What should be the width of each slit in order to obtain 10 maxima of the double slit pattern within the central maxima of the single slit pattern?
U [OD 2016]
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Solution:
Ans. (i)
$$ \begin{aligned} \beta & =\frac{\lambda D}{d} \ & =\frac{500 \times 10^{-9} \times 1}{10^{-3}} \ & =0.5 \mathrm{~mm} \text { or } 5 \times 10^{-4} \mathrm{~m} \end{aligned} $$
(ii)
$$ \begin{aligned} \beta_{0} & =\frac{2 \lambda D}{\alpha}=10 \beta \ \alpha & =\frac{2 \times 500 \times 10^{-9} \times 1}{10 \times 5 \times 10^{-4}} \ \alpha & =2 \times 10^{-4} \mathrm{~m} \text { or } 0.2 \mathrm{~mm} 1 / 2 \end{aligned} $$
[CBSE Marking Scheme 2016]
