A fringe width of 6 mm was produced for two slits separated by 1 mm apart. The screen is placed 10 m away. The wavelength of light used is 'x' nm. The value of 'x' to the nearest integer is ____________.
Answer (integer)
600
Solution
$\beta$ = 6 mm, d = 1 mm, D = 10 m<br><br>$\lambda$ = ?<br><br>We know, $\beta = {{\lambda D} \over d}$<br><br>6 $\times$ 10<sup>$-$3</sup> = ${{\lambda \times 10} \over {1 \times {{10}^{ - 3}}}}$<br><br>$\therefore$ $\lambda$ = ${{6 \times {{10}^{ - 3}} \times 1 \times {{10}^{ - 3}}} \over {10}}$<br><br>$\lambda$ = 600 $\times$ 10<sup>$-$9</sup> m<br><br>$\therefore$ $\lambda$ = 600 nm
About this question
Subject: Physics · Chapter: Optics · Topic: Reflection and Mirrors
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