Question

A double-slit is illuminated with monochromatic light in the air and it turns out that the first order minima occur at angles ∓35.09 ° relative to the reference line from the double slit to central max. If the double slit is lowered into a transparent liquid and illuminated with the same light, the first order minima occur at the angles ± 19.36 °. Decide the refractive index of the liquid?

Answer 1

Equation for first order minimum for interference

sin=/2d

=angular width

=wavelength

d=distance between slits

that is sin (as d is same in 2 cases)

for first case =35.09 and let wavelength=

sin35.09     ...................(1)

in second case new wavelength

'=/n where n= refractive index of the liquid (medium)

=19.36

Therefore

sin19.36    /n....(2)

dividing equation (1) by (2)

sin35.09/sin19.36=/(/n)=n

n=sin35.09/sin19.36=1.734=refractive index of liquid