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Wave Optics — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsWave Optics3 Marks
Question
In a biprism experiment, a source of light having wavelength $6500 A$ is replaced by a source of light having wavelength $5500 A$. Calculate the change in the fringe width, if the screen is at a distance of $1 m$ from the sources which are $1\ mm$ apart.
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Answer

Data : $\lambda_1=6500 \mathring A =6.5 \times 10^{-7} m$,
$\lambda_2=5500 \mathring A =5.5 \times 10^{-7} m ,$
$D=1 m ,$
$d=1\ mm =1 \times 10^{-3} m$
Fringe width, $W=\frac{\lambda D}{d}$
$\therefore W_1=\frac{\lambda_1 D}{d}$ and $W_2=\frac{\lambda_2 D}{d}$
Since $\lambda_1>\lambda_2, W_1>W_2$.
$\therefore W_1-W_2 =\left(\lambda_1-\lambda_2\right) \frac{D}{d}$
$ =(6.5-5.5) \times 10^{-7} \times \frac{1}{1 \times 10^{-3}}$
$ =1 \times 10^{-4} m =0.1\ mm$
The fringe width decreases by $1 \times 10^{-4} m =0.1\ mm$.

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