Question

need help. please answer #1 and 2

1. You are looking, from directly above, at a goldfish at the bottom of a 54.0-cm fish tank filled with water. How far away does the fish look to you? The index of refraction of water is 1.34. 2. A beam of light is incident, from air, on a flat surface of glass (n=1.50). The refracted ray inside the glass is found to make an angel of 26.7° with the direction of the incident ray. Calculate the angle of incident.

Answer 1

1) given

lamda = 650 nm = 650*10^-9 m
distance between central nright fringe to 1st bright fringe, y1_max = 7 cm

= 0.07 m

line spacing on the grating, d = 1 mm/100

= 10^-2 mm

= 10^-5 m

let R is the distance from grating to screen.

we know, y1_max = lamda*R/d

==> R = y1_max*d/lamda

= 0.07*10^-5/(650*10^-9)

= 1.08 m <<<<<<<<<<<<-----------------Answer

1) given
real depth at which the fish swims, d = 54.0 cm
refractive index, n = 1.34

apparent depeth of the fish, d' = real depth/n

= 54/1.34

= 40.3 cm <<<<<<<<<-----------------Answer

2)

given
n1 = 1 (for air)
n2 = 1.5 (for glass)
angle of refraction, theta_r = theta_i - 26.7

using Snell's law, sin(theta_i)/sin(theta_r) = n2/n1

sin(theta_i)/sin(theta_i - 26.7) = 1.5/1

let theta_i = x

sin(x)/sin(x - 26.7) = 1.5

sin(x) = 1.5*sin(x - 26.7)

solving above equation we get, x = 63.2 degrees

so, angle of incidence, theta_i = 63.2 degrees <<<<<<<<<<<---------------Answer