A mixture of light, consisting of wavelength 590 nm 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 4th  bright fringe of the unknown light. From this data, the wavelength of the unknown light is

  • Option 1)

    393.4 nm

  • Option 2)

    885.0 nm

  • Option 3)

    442.5 nm

  • Option 4)

    776.8 nm

 

Answers (1)

As we learnt in

Fringe Width -

\beta = \frac{\lambda D}{d}
 

- wherein

\beta = y_{n+1}-y_{n}

y_{n+1}= Distance of\left ( n+1 \right )^{th}

Maxima= \left ( n+1 \right )\frac{\lambda D}{d}

y_{n}=Distance of n^{th}

 maxima = \frac{n\lambda D}{d}

 

 3rd bright fringe of known light coincides with the 4th bright fringe of the unknown light.

\therefore\ \; \frac{3(590)\Delta}{d}=\frac{4\lambda \Delta}{d}

\Rightarrow\ \; \lambda=\frac{3}{4}\times590=442.5\ nm

Correct option is 3.

 


Option 1)

393.4 nm

This is an incorrect option.

Option 2)

885.0 nm

This is an incorrect option.

Option 3)

442.5 nm

This is the correct option.

Option 4)

776.8 nm

This is an incorrect option.

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