MHT CET · Physics · Wave Optics
The Young's double slit experiment, angular width of fringes is 0.20 radians for sodium light of wavelength \(5890 Å\). If the complete system is dipped in water, then the angular width of fringes will be
\(\text { [Refractive index of water } \left.=\frac{4}{3}\right]\)
- A \(0.30^{\circ}\)
- B \(0.15^{\circ}\)
- C \(0.11^{\circ}\)
- D \(0.22^{\circ}\)
Answer & Solution
Correct Answer
(B) \(0.15^{\circ}\)
Step-by-step Solution
Detailed explanation
Angular fringe width is given by
\(\beta=\frac{\lambda}{d}\)
\(\beta=0.2\) radians \(=\frac{\lambda}{d}\)
If dipped in water
\(\beta^{\prime}=\frac{\lambda^{\prime}}{d}\)
where \(\lambda^{\prime}=\frac{\lambda}{\mu}\) for water, \(\mu=\frac{4}{3}\)
Taking ratio of \(\mathrm{eq}^{\mathrm{n}}\) (1) and \(\mathrm{eq}^{\mathrm{n}}\) (2)
\(\begin{aligned}
& \frac{0.2}{\beta^{\prime}}=\frac{\left(\frac{\lambda}{d}\right)}{\left(\frac{3 \lambda}{4 d}\right)} \\
& \Rightarrow \beta^{\prime}=0.2 \times \frac{3}{4} \text { radians }=0.15 \text { radians }
\end{aligned}\)
\(\beta=\frac{\lambda}{d}\)
\(\beta=0.2\) radians \(=\frac{\lambda}{d}\)
If dipped in water
\(\beta^{\prime}=\frac{\lambda^{\prime}}{d}\)
where \(\lambda^{\prime}=\frac{\lambda}{\mu}\) for water, \(\mu=\frac{4}{3}\)
Taking ratio of \(\mathrm{eq}^{\mathrm{n}}\) (1) and \(\mathrm{eq}^{\mathrm{n}}\) (2)
\(\begin{aligned}
& \frac{0.2}{\beta^{\prime}}=\frac{\left(\frac{\lambda}{d}\right)}{\left(\frac{3 \lambda}{4 d}\right)} \\
& \Rightarrow \beta^{\prime}=0.2 \times \frac{3}{4} \text { radians }=0.15 \text { radians }
\end{aligned}\)
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