CBSE Class 10 Answered

Please answer questions 22, 23 and and 24.

Asked by guptaabhinav0809 | 23 Nov, 2018, 06:21: PM

Expert Answer

Figure shows the incident ray is getting deviated as RS in a prism due to refraction.

In the quadrilateral AQNR, two of the angles (at the vertices Q and R) are right angles. Therefore, the sum of the other angles of the
quadrilateral is 180º.
∠A + ∠QNR = 180º
From the triangle QNR, r1 + r2 + ∠QNR = 180º
Comparing these two equations, we get r1 + r2 = A ..........................(1)
The total deviation δ is the sum of deviations at the two faces, δ = (i – r1 ) + (e – r2 ) ; that is, δ = i + e – A ...............(2)

Thus, the angle of deviation depends on the angle of incidence. A plot between the angle of deviation and angle of incidence is shown in
figure given below

We can see that, in general, any given value of δ, except for i = e, corresponds to two values i and hence of e.

This, in fact, is expected from the symmetry of i and e in Eq. (2), i.e., δ remains the same if i and e are interchanged.

Physically, this is related to the fact that the path of ray can be traced back, resulting in the same angle of deviation.

At the minimum deviation Dm, the refracted ray inside the prism becomes parallel to its base.

We have δ = Dm, i = e which implies r1 = r2. Hence equation (1) gives 2r = A or r = 2A ..................(3)

In the same way, Eq. (2) gives Dm = 2i – A, or i = (A + Dm)/2 ..........................(3)

The refractive index μ of the prism is

begin mathsize 12px style mu space equals space fraction numerator sin space i over denominator sin space r end fraction space equals space fraction numerator sin open square brackets A plus begin display style D subscript m over 2 end style close square brackets over denominator sin space begin display style A over 2 end style end fraction end style

Thus angle of minimum deviation is related to angle of prism, and refractive index of material

Answered by Thiyagarajan K | 24 Nov, 2018, 09:15: PM

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CBSE Class 10 Answered

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