In a Young's double slit experiment two slits are separated by 2 mm and the screen is placed one meter away. When a light of wavelength 500 nm is used, the fringe separation will be :
Solution
Fringe width ($\beta$) = ${{\lambda D} \over d}$<br><br>d = 2 $\times$ 10<sup>$-$3</sup> m<br><br>$\lambda$ = 500 $\times$ 10<sup>$-$9</sup> m<br><br>D = 1 m<br><br>Now<br><br>$\beta$ = ${{500 \times {{10}^{ - 9}} \times 1} \over {2 \times {{10}^{ - 3}}}}$<br><br>$\beta$ = ${5 \over 2} \times {10^{ - 4}}$<br><br>$\beta$ = 2.5 $\times$ 10<sup>$-$4</sup><br><br>$\beta$ = 0.25 mm
About this question
Subject: Physics · Chapter: Optics · Topic: Reflection and Mirrors
This question is part of PrepWiser's free JEE Main question bank. 197 more solved questions on Optics are available — start with the harder ones if your accuracy is >70%.