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Minimum deviation of a thin glass prism when it is immersed in water

The same question, asked 4 times: 15 May 2023 S2, 14 May 2024 S1, 19 Apr 2025 S1, 11 Apr 2026 S2. Every wording below, oldest first.

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  1. Question 1

    15 May 2023 S2

    38
    The angle of minimum deviation produced by a thin prism in air is δ1\delta_1. If it is immersed in water the angle of minimum deviation is (ag=3/2, aw=4/3{}^a g = 3/2,\, {}^a w = 4/3)

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  2. Question 2

    14 May 2024 S1

    7
    The angle of deviation produced by a thin prism when placed in air is 'δ1\delta_1' and that when immersed in water is 'δ2\delta_2'. The refractive index of glass and water are 32\dfrac{3}{2} and 43\dfrac{4}{3} respectively. The ratio δ1:δ2\delta_1:\delta_2 is

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  3. Question 3

    19 Apr 2025 S1

    25
    For a thin prism, δ1\delta_{1} is the angle of deviation produced, when prism is placed in air. When the prism is immersed in water, the angle of deviation produced is δ2\delta_{2}. Given  aμg=32\ _{a}\mu_{g}=\frac{3}{2} and  aμw=43\ _{a}\mu_{w}=\frac{4}{3}. The ratio δ2:δ1\delta_{2}:\delta_{1} is

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  4. Question 4

    11 Apr 2026 S2

    140
    A thin prism in air produces the angle of minimum deviation δ\delta. If the prism is immersed in water, the angle of minimum deviation for the same ray is (refractive index of water is 4/34/3 and that of glass prism is 3/23/2)

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