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A ray of white light traveling through fused quartz is incident at a quartz–air interface at angleθ1.Assume that the index of refraction of quartz is n=1.456at the red end of the visible range andn=1.470 at the blue end.Ifθ1 is(a) θ1=42.00°, (b),θ1=43.10°and (c),θ1=44.00°is the refracted light white, white dominated by the red end of the visible range, or white dominated by the blue end of the visible range, or is there no refracted light?

Short Answer

Expert verified

(a) The refracted light for the given incidence angle42.00° will be white light.

(b) The refracted light for the given incidence angle 43.10°will be white dominated by the red end of the visible end.

(c) There is no refracted light for the given incidence angle.44.00°

Step by step solution

01

The given data:

  • The index of refraction of quartz at the red end of the visible range is,n=1.456.
  • The index of refraction of quartz at the blue end of the visible range is,n=1.470 .
02

Understanding the concept of refraction: 

Use the concept of refraction of light. Calculate the critical angle for the quartz at both the red end of the visible range and at the blue end. So, you can compare the given incidence angle with the critical angle and can determine the nature of refracted light.

Formula:

The critical angle of a refracted light with the incident light,

θc=sin−1(n2n1)….. (1)

03

a) Calculation of the refracted light for angle  : 42° 

Here, the ray of white light is incident at a quartz-air interface. Therefore,

The index of refraction of air, n2=1

The index of refraction, n1=n

Thus, using this value in equation (1), you can get the equation of critical angle as follows:

θc=sin−1(1n) ….. (2)

Thus, the critical angle for the red end of the visible range is given using the given data in equation (2) as:

(θc)red=sin−1(11.456)=43.38°

Similarly, the critical angle for the blue end of the visible range is given using the given data in equation (a) as:

(θc)blue=sin−1(11.470)=42.86°

Now, considering the incidence angleθ1as42.00° , you get that the incident angle is less than both critical angle for red end and critical angle for blue end.

So, all types of rays can be refracted. Hence, the refracted light will be white light.

04

(b) Calculation of the refracted light for angle  :43.10°

Now, considering the incidence angleθ1as 43.10°, you get that the incident angle is greater than the critical angle for the blue end of the visible range and less than the red end of the visible range.

Hence, the refracted light will be white-dominated by the red end of the visible end.

05

(c) Calculation of the refracted light for angle  :

Now, considering incidence angleθ1as 44.00°, you get that the incident angle is less than both the critical angle for the red end and the critical angle for the blue end.

Hence, no visible light can be refracted, and you can say that there is no refracted light.

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