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Hard Interference: Young's double-slit experiment and expression for fringe width, coherent sources and sustained interference of light MCQs for JEE

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Part of Optics in JEE Main Physics.

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Q1PhysicsUnit 16: Optics

Two coherent monochromatic light beams of intensities

I

and

4 I

are superposed. The maximum and minimum possible intensities in the resulting beam are:

Q2PhysicsUnit 16: Optics

Two independent monochromatic sodium lamps can not produce interference because

Q3PhysicsUnit 16: Optics

Assertion Ratio of maximum intensity and minimum intensity in interference is 25 1. Hence amplitude ratio of two waves should be 3: 2 Reason

\frac{\boldsymbol{I}_{m a x}}{\boldsymbol{I}_{m i n}}=\left(\frac{\boldsymbol{A}_{1}+\boldsymbol{A}_{2}}{\boldsymbol{A}_{1}-\boldsymbol{A}_{2}}\right)^{2}

Q4PhysicsUnit 16: Optics

A mixture of light, consisting of wavelength 590nm and an unknown wavelength, illuminates Young's double slit and gives rise to two overlapping interference patterns on the screen. The central maximum of both lights coincide. Further, it is observed that the third bright fringe of known light coincides with the 4 th bright fringe of the unknown light. From this data, the wavelength of the unknown light is

Q5PhysicsUnit 16: Optics

Calculate the number of fringes.

Q6PhysicsUnit 16: Optics

The maximum number of possible interference maxima, for slit separation equal to twice the wavelength,in Young's double slit experiment is :

Q7PhysicsUnit 16: Optics

In a Young's double slit experiment, constructive interference is produced at a certain point

P .

The intensities of light at

\boldsymbol{P}

due to the individual sources are 4 and 9 units. The resultant intensity at point

\boldsymbol{P}

will be-

Q8PhysicsUnit 16: Optics

Find the fringe width of the fringe pattern

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