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Light from a source emitting two wavelengths \mathrm{ \lambda_1}  and \mathrm{ \lambda_2} is allowed to fall on Young's double - slit apparatus after filtering one of the wavelengths. The position of interference maxima is noted. When the filter is removed both the wavelengths are incident and it is found that maximum intensity is produced where the four maxima occurred previously. If the other wavelength is filtered, at the same location the third maxima is found. Find the ratio of wavelengths:

Option: 1

\frac{2}{3}


Option: 2

\frac{3}{5}


Option: 3

\frac{3}{4}


Option: 4

\frac{4}{5}


Answers (1)

best_answer

The maxima is obtained whenever \mathrm{\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}}}

For first wavelength, \mathrm{\left(x_4\right)_{\lambda_1}=\frac{4 \lambda_1 \mathrm{D}}{\mathrm{d}}}

For second wavelength, \mathrm{\left(x_x\right)_{\lambda_2}=\frac{3 \lambda_2 D}{d}}

\mathrm{ \begin{aligned} & \because \quad\left(\mathrm{x}_4\right)_{\lambda_1}=\left(\mathrm{x}_3\right)_{\lambda_2} \\\\ & \therefore \quad \frac{4 \lambda_1 \mathrm{D}}{\mathrm{d}}=\frac{3 \lambda_2 \mathrm{D}}{\mathrm{d}} \Rightarrow \frac{\lambda_1}{\lambda_2}=\frac{3}{4} \end{aligned} }

Posted by

Sanket Gandhi

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