Question

A single slit diffraction pattern is obtained on a screen using 600 nm light when the width of the central (principal) maximum is 0.30 mm. The width of the slit is reduced by 50%. and then illuminated by 450 nm light. Width, in mm, of the central maximum in the second case is;

• A. 0.45
• B. 0.4
• C. 0.25
• D. 0.2

Answer 1

The Correct Option is A

Width of central maxima, $\beta = \frac{2 \lambda D}{a}$
Given , $ \lambda_1 = 600 \, nm, \beta_1 = 0.30 \, mm, a_1 = a, D_1 =D$
$\lambda_{2} =450 nm , a_{2} = \frac{a_{1}}{2} =\frac{a}{2}, D_{2} =D_{1} , \beta_{2} =? $
$\therefore \, \frac{\beta_{1}}{\beta_{2} } = \frac{\lambda _{1}}{\lambda _{2}} \times\frac{D_{1}}{D_{2}} \times \frac{a_{2}}{a_{1}} = \frac{600}{450} \times \frac{D}{D} \times \frac{a / 2}{a} $
or, $\frac{0.30 \: mm}{\beta_{2}} = \frac{600}{900} = \frac{2}{3} $
or, $\beta_{2} = 0.45 mm$