In a Young’s double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away.  A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen.  The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is : Option 1)  1.56 mm   Option 2)  7.8 mm   Option 3)  9.75 mm   Option 4) 15.6 mm

As we learnt in

Young Double Slit Experiment -

- wherein

Distance of a point on screen from central maxima

Path difference at that point

Ref y be the distance from the central maximum to the point where the bright fringes due to both the wavelength coincides

For ,

for  ,

i.e. with respect to central maximum 4th bright fringe of  coincides with 5th bright fringe of

or     y = 7.8 mm

Correct option is 2.

Option 1)

1.56 mm

This is an incorrect option.

Option 2)

7.8 mm

This is the correct option.

Option 3)

9.75 mm

This is an incorrect option.

Option 4)

15.6 mm

This is an incorrect option.

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