JEE MainPhysicsWave OpticsMCQ+4 / −1
The width of one of the two slits in a Young's double slit experiment is 4 times that of the other slit. The ratio of the maximum of the minimum intensity in the interference pattern is:
- A
- B
- C
- D
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Correct answer: 25:9, NONE OF THE OPTIONS
- Relate slit width to amplitude
In Young's double slit experiment, the amplitude contributed by a slit is proportional to its width.
Let the smaller slit have width . Then the larger slit has width .
So the corresponding amplitudes are in the ratio
- Write expressions for maximum and minimum intensity
If the amplitudes of the two waves are and , then
- Maximum intensity:
- Minimum intensity:
Taking and ,
- Find the ratio
- Compare with the given options
The correct ratio is
This is not present in the listed options.
So none of the options is correct.
- Compare with stored correct answer
The stored correct answer is (), but the derived answer is .
Hence, the stored answer appears to be incorrect.
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