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Question Bank [ 2022 ] — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsQuestion Bank [ 2022 ]3 Marks
Question
Answer in brief:
The distance between two consecutive bright fringes in a biprism experiment using the light of wavelength $6000 \mathring A$ is $0.32\ mm$ by how much will the distance change if light of wavelength $4800 \mathring A$ is used?
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Answer

Data: $\lambda_1=6000 Å =6 \times 10^{-7} m , \lambda_2=4800 Å=4.8 \times 10^{-7} m$, $W _1=0.32 mm =3.2 \times 10^{-4} m$
Distance between consecutive bright fringes,
$W =\frac{\lambda D }{ d }$
For $\lambda_1, W _1=\frac{\lambda_1 D }{ d }$ and
For $\lambda_2, W _2=\frac{\lambda_2 D }{ d }$ and
$ \frac{ W _2}{ W _1}=\frac{\lambda_2 D / d }{\lambda_1 D / d }=\frac{\lambda_2}{\lambda_1}$
$\therefore W _2=\left(\frac{\lambda_2}{\lambda_1}\right) W _1=\left(\frac{4.8 \times 10^{-7}}{6 \times 10^{-7}}\right)\left(3.2 \times 10^{-4}\right)$
$=(0.8)\left(3.2 \times 10^{-4}\right) m$
$=2.56 \times 10^{-4} m$
$\therefore \triangle W = W _1- W _2 $
$=3.2 \times 10^{-4} m -2.56 \times 10^{-4} m$
$=0.64 \times 10^{-4} m$
$=6.4 \times 10^{-5} m$
$= 0 . 0 6 4\ mm $

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