CBSE BOARD XII, asked by shelicesrahRohe, 1 year ago

Two monochromatic rays of light are incident normally on the face AB of an isosceles right-angled prism ABC. The refractive indices of the glass prism for the two rays '1' and '2' are respectively 1.35 and 1.45. Trace the path of these rays after entering the prism.

Answers

Answered by rishilaugh
15
Critical angle of ray 1: sinc1=1 μ1=11.35⇒c1=sin-111.35=47.73° Similarly, critical angle of ray 2: sinc2=1μ2=11.45⇒c2=sin-111.45=43.6° Both the rays will fall on the side AC with angle of incidence (i) equal to 45°. Critical angle of ray 1 is greater than that of i. Hence, it will emerge from the prism, as shown in the figure. Critical angle of ray 2 is less than that of i. Hence, it will be internally reflected, as shown in the figure.
Answered by nalinsingh
22

Hey !!

At plane AC, the incident angle for ray 1 and ray 2 = 45°

Let critical angle for total internal reflection for ray 1 = C₁

                                     1.35 = 1/sinC₁

=>            sin C₁ = 1/1.35

                          = 0.74

Hence,                   C₁ > 45°         ( sin 45° = 0.707 )

Let critical angle for total internal reflection for ray 2 = C₂

                 1.45 = 1/sin C₂

=>             sin C₂ = 1/1.45 = 0.689

 Hence,

                        C₂ < 45°     ( sin 45° = 0.707 )

As in case of ray 1, incident angle is less than critical angle, it would emerge out from AC. In the figure of the ray 1 is shown.

In case of ray 2, incident angle is greater than critical angle, it would get total internal reflection at AC and emerge from BC. In the figure path 2 is shown.


GOOD LUCK !!


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