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Q.Read the following passage carefully and answer the questions given below: A prism is an optical medium bounded by three refracting plane surfaces. A ray of light suffers successive refraction on passing through its two surfaces and deviates by a certain angle from its original path. The refractive index of the material of the prism is given by: n=sin⁡(A+δm2)/sin⁡A2n = \sin\left(\dfrac{A+\delta_m}{2}\right)\big/\sin\dfrac{A}{2}. If the angle of incidence on the second surface is greater than an angle called the critical angle, the ray will not be refracted from the second surface and is totally internally reflected. (क/a) [1 mark, MCQ] The critical angle for glass is θ₁ and that for water is θ₂. The critical angle for the glass-water surface would be (given: refractive index of glass w.r.t. air ang=1.5_a n_g=1.5 and refractive index of water w.r.t. air anw=1.33_a n_w=1.33): options —

(i) less than θ₂,
(ii) between θ₁ and θ₂,
(iii) greater than θ₂,
(iv) less than θ₁. (ख/b) [1 mark] When a ray of light of wavelength λ and frequency ν is refracted into a denser medium, what changes happen in λ and ν? (ग/c) [2 marks]
A right-angled glass prism ABC: vertex A at top with angle A=60 degrees, vertex B bottom-left with a right angle 90 degrees (small square), — Class 12 Physics question
Figure
A ray of light is incident normally on a prism ABC of refractive index 2\sqrt{2}, as shown in the figure. After it strikes face AC, what will be the path of this ray?
Uttarakhand UbseUttarakhand Board Intermediate (Class 12) 2025Subjective· 4mImportance★★★★★
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Critical angle rises as the refractive-index mismatch across the interface falls, wavelength shortens (frequency stays fixed) on entering a denser medium, and in the right-angled prism the ray totally internally reflects once (at AC) before finally emerging through the base BC.

(क/a) Critical angle for glass–water interface [1 mark]: For total internal reflection between two media of refractive indices n1n_1 (denser) and n2n_2 (rarer), sin⁡θc=n2/n1\sin\theta_c = n_2/n_1.

For glass–air: sin⁡θ1=1ang=11.5=0.667  ⟹  θ1≈41.8∘\sin\theta_1 = \dfrac{1}{{}_an_g} = \dfrac{1}{1.5} = 0.667 \implies \theta_1 \approx 41.8^\circ.

For water–air: sin⁡θ2=1anw=11.33=0.752  ⟹  θ2≈48.8∘\sin\theta_2 = \dfrac{1}{{}_an_w} = \dfrac{1}{1.33} = 0.752 \implies \theta_2 \approx 48.8^\circ.

For glass–water (going from denser glass into rarer water): sin⁡θc=anwang=1.331.5=0.887  ⟹  θc≈62.5∘\sin\theta_c = \dfrac{{}_an_w}{{}_an_g} = \dfrac{1.33}{1.5} = 0.887 \implies \theta_c \approx 62.5^\circ.

Since θc(62.5∘)>θ2(48.8∘)>θ1(41.8∘)\theta_c(62.5^\circ) > \theta_2(48.8^\circ) > \theta_1(41.8^\circ), the glass–water critical angle is greater than θ₂.

(ख/b) Refraction into a denser medium [1 mark]: When light of wavelength λ\lambda and frequency ν\nu enters a denser medium, its speed decreases. Since the frequency ν\nu is determined by the source and does not change on refraction, and v=νλv = \nu\lambda, the decrease in speed vv must be accompanied by a decrease in wavelength λ\lambda. So: wavelength decreases, frequency remains unchanged.

(ग/c) Path of the ray in the prism [2 marks]: The prism ABC has ∠A=60∘\angle A = 60^\circ and a right angle at B, so ∠C=180∘−60∘−90∘=30∘\angle C = 180^\circ-60^\circ-90^\circ = 30^\circ.

The ray enters face AB normally (along the horizontal, perpendicular to the vertical face AB), so it passes straight through without bending and travels horizontally inside the prism until it strikes face AC.

Because the ray travels parallel to the base BC, and side AC makes an angle of ∠C=30∘\angle C = 30^\circ with BC, by simple geometry the ray also makes an angle of 30∘30^\circ with the surface AC at the point where it strikes it. So the angle of incidence at AC, measured from the normal to AC, is 90∘−30∘=60∘90^\circ-30^\circ = 60^\circ.

The critical angle for the prism material (refractive index n=2n=\sqrt2) is:

sin⁡θc=1n=12  ⟹  θc=45∘\sin\theta_c = \dfrac1n = \dfrac{1}{\sqrt2} \implies \theta_c = 45^\circ

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