JEE Mains · Physics · STD 12 - Wave optics
- JEE Mains 2025
- Easy
- MCQ
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion-(A) : If Young's double slit experiment is performed in an optically denser medium than air, then the consecutive fringes come closer.
Reason-(R) : The speed of light reduces in an optically denser medium than air while its frequency does not change.
In the light of the above statements, choose the most appropriate answer from the options given below :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) Both \((\mathbf{A})\) and \((\mathbf{R})\) are true and \((\mathbf{R})\) is the correct explanation of \((\mathbf{A})\)
Step-by-step Solution
Detailed explanation
\(\beta(\text { fringe width })=\frac{\lambda D}{d}\)
In denser medium, \(\lambda \downarrow \Rightarrow \beta \downarrow\) \(\Rightarrow\) fringe come closer
Also, \(\mu=\frac{\mathrm{c}}{\mathrm{V}} \Rightarrow \mathrm{V}=\frac{\mathrm{c}}{\mu}\)
Frequency remains same,
\(\Rightarrow \mu=\frac{\lambda_{\mathrm{vac} .} \mathrm{f}}{\lambda_{\mathrm{med}} \mathrm{f}} \Rightarrow \lambda_{\mathrm{med}}=\frac{\lambda_{\mathrm{vac} .}}{\mu}\)
In denser medium, \(\lambda \downarrow \Rightarrow \beta \downarrow\) \(\Rightarrow\) fringe come closer
Also, \(\mu=\frac{\mathrm{c}}{\mathrm{V}} \Rightarrow \mathrm{V}=\frac{\mathrm{c}}{\mu}\)
Frequency remains same,
\(\Rightarrow \mu=\frac{\lambda_{\mathrm{vac} .} \mathrm{f}}{\lambda_{\mathrm{med}} \mathrm{f}} \Rightarrow \lambda_{\mathrm{med}}=\frac{\lambda_{\mathrm{vac} .}}{\mu}\)
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