Question

In Young's double slit experiment the fringe width with light of wavelength ${6000}^{\circ }A$ is found to be $4.0mm$

• A. fringe width if light of wavelength $4800{}^{\circ }A$ is used is $3.2mm$
• B. distance between central maxima and first minimum is $3.2mm$
• C. central maxima is brighter than first maxima
• D. brightness of first minimum is same as that of second minimum

Answer 1

Answer: (A)

fringe width if light of wavelength $4800{}^{\circ }A$ is used is $3.2mm$

Given,

$\lambda =6000A^{\circ }=6\times {10}^{-7}m\beta =4mm=4\times {10}^{-3}mm$

We know that fringe width $\beta$ is

$\beta =\frac{\lambda D}{d}$

Here $\beta \propto \lambda$

Now,

$\frac{{\beta }_1}{{\beta }_2}=\frac{{\lambda }_1}{{\lambda }_2}\frac{4\times {10}^{-3}}{{\beta }_2}=\frac{6\times {10}^{-7}}{4.8\times {10}^{-7}}{\beta }_2=\frac{4\times {10}^{-3}}{1.25}=3.2mm$

Correct option is A.