JEE Mains · Physics · STD 12 - Wave optics
- JEE Mains 2021
- Hard
- Numerical
The width of one of the two slits in a Young's double slit experiment is three times the other slit. If the amplitude of the light coming from a slit is proportional to the slit-width, the ratio of minimum to maximum intensity in the interference pattern is \(x: 4\) where \(x\) is ..... .
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Answer & Solution
Correct Answer
(A) \(1\)
Step-by-step Solution
Detailed explanation
Given amplitude \(\propto\) slit width
Also intensity \(\propto(\text { Amplitude })^{2} \propto\) (Slit width \()^{2}\)
\(\frac{{I}_{1}}{{I}_{2}}=\left(\frac{3}{1}\right)^{2}=9 \Rightarrow {I}_{1}=9 {I}_{2}\)
\(\frac{{I}_{\min }}{{I}_{\max }}=\left(\frac{\sqrt{{I}_{1}}-\sqrt{{I}_{2}}}{\sqrt{{I}_{1}}+\sqrt{{I}_{2}}}\right)^{2}=\left(\frac{3-1}{3+1}\right)^{2}=\frac{1}{4}=\frac{{x}}{4}\)
\(\Rightarrow {x}=1.00\)
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