Question

A flint glass plate (n = 1.66) rests on the bottom of an aquarium tank. The plate is 10.00 cm thick (vertical dimension) and covered with water (n = 1.33) to a depth of 12.3 cm. Calculate the apparent thickness of the plate as viewed from above the water. (Assume nearly normal incidence of light rays.)

Answer 1

For nearly normal incidence of light rays,

Refractive index of the flint glass plate = real base / apparent base

                                      Apparent base = real base / Refractive index of the flint glass plate

                                                              = 10.00 cm / 1.66

                                                              = 6.0241 cm

When it is viewed through water, the refractive index of water is

     refractive index of water = real depth of the base / Apparent depth of the base

Apparent depth of the base = real depth of the base / refractive index of water

                                            = (12.3 cm + 6.0241 cm) / 1.33

                                            = 13.778 cm

Consider only the top surface of the flint glass.

Refractive index of water = real depth / apparent depth

                Apparent depth = real depth / refractive index of water

                                          = 12.3 cm / 1.33

                                           = 9.2481 cm

Thus, the apparent thickness of the plate as viewed from above the water is

y = Apparent depth of the base - Apparent depth

= 13.778 cm - 9.2481 cm

   = 4.5299 cm

Rounding off to three significant figures, the apparent thickness of the plate as viewed from above the water is 4.53 cm.