Question

A glass sheet measuring 14.5 cm × 24.0 cm is covered by a very thin opaque coating. In the middle of this sheet is a thin, straight scratch 0.00140 mm thick. The sheet is totally immersed beneath the surface of a liquid having an index of refraction of 1.44. Monochromatic light strikes the sheet perpendicular to its surface and passes through the scratch. A screen is placed under water a distance 37.0 cm away from the sheet and parallel to it. You observe that the first dark fringes on either side of the central bright fringe on this screen are 22.4 cm apart. What is the wavelength of the light in air?

Answer 1

thickness of scratch, b = 0.00140 mm = 1.4*10^-6 m

distance of screen, D = 37 cm = 0.37 m

refractive index of liquid = 1.44

For first minima, $sin \theta = \frac{\lambda_l }{b}$

Also $sin \theta = \frac{0.224/2}{\sqrt{0.37^2 + (0.224/2)^2}} = 0.28972$

So, $0.28972 = \frac{\lambda_l }{1.4*10^{-6} }$

=> $\lambda _l = 0.4056 * 10^{-6} m$

the wavelength of the light in air = wavelength in liquid * refractive index = 0.4056*10^-6 * 1.44 = 0.584*10^-6 m = 584 nm