An aquarium filled with water (nwater = 1.33) has flat glass sides whose index o

Question

An aquarium filled with water (nwater = 1.33) has flat glass sides whose index of refraction is 1.58. A beam of light from outside the aquarium (n air= 1.00) strikes the glass at an angle of 30.0 degrees as measured from outside of the glass to the incident light ray.

What is the angle of refraction when:

A) The light rays enters the glass from air?

B) the light ray enter the water from the glass?

C) what is the critical angle from the glass to the water?

Please give a detailed ray diagram of the situations

Explanation / Answer

(A) Here, the angle of refraction for the light enters the glass from air is determined from the law of refraction -

n1 sinx1 = n2 sin x2

n1 =1 (since the light is in air)
x1 = 30.0 deg
n2 =1.58

therefore -

sin x2 = 1(sin 30.0)/1.58 = 0.862

=> x2 = arcsin(0.862) = 59.5 deg

(B) when this enters water (n=1.33), we have

1.33 sin(x water)= 1.58 sin 59.5

=> sin(x water) = 1.58 sin 59.5/1.33 = 1.02

Please note that this value is greater than 1. So, beam of light will not enter the water rather it will be reflectled.

(C) For determining the critical angle from glass to water -

we will have to consider, xwater = 90 deg.

put these values in the above expression -

1.33 sin(90)= 1.58 sin (theta)c

=> sin (theta)c= 1.33 / 1.58 = 0.842

=> (theta)c= 57.3 deg.

So, the critical angle from glass to the water = 57.3 deg.

Draw the ray diagram with the help of Snell's Law.

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