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Geometrical Optics question

2025 · 28 Jan · Shift 1 · Q66
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  5. /2025 · 28 Jan · Shift 1 · Q66

Geometrical Optics question

2025 · 28 Jan · Shift 1 · Q66

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
A thin prism P1\mathrm{P}_1P1​ with angle 4∘4^{\circ}4∘ made of glass having refractive index 1.54 , is combined with another thin prism P2\mathrm{P}_2P2​ made of glass having refractive index 1.72 to get dispersion without deviation. The angle of the prism P2\mathrm{P}_2P2​ in degrees is
  1. A
    1.5
  2. B
    16/3
  3. C
    3
  4. D
    4
View written solutionFree

Correct answer: C

  1. Condition for dispersion without deviation

For two thin prisms combined to produce dispersion without net deviation, the total mean deviation must be zero.

For a thin prism,

δ=(μ−1)A\delta = (\mu-1)Aδ=(μ−1)A

where μ\muμ is refractive index and AAA is prism angle.

So for the two prisms placed oppositely,

(μ1−1)A1=(μ2−1)A2(\mu_1-1)A_1 = (\mu_2-1)A_2(μ1​−1)A1​=(μ2​−1)A2​
  1. Substitute given values

For prism P1P_1P1​:

μ1=1.54,A1=4∘\mu_1 = 1.54, \quad A_1 = 4^\circμ1​=1.54,A1​=4∘

For prism P2P_2P2​:

μ2=1.72\mu_2 = 1.72μ2​=1.72

Thus,

(1.54−1)(4)=(1.72−1)A2(1.54-1)(4) = (1.72-1)A_2(1.54−1)(4)=(1.72−1)A2​ 0.54×4=0.72A20.54 \times 4 = 0.72 A_20.54×4=0.72A2​ 2.16=0.72A22.16 = 0.72 A_22.16=0.72A2​ A2=2.160.72=3∘A_2 = \frac{2.16}{0.72} = 3^\circA2​=0.722.16​=3∘
  1. Match with options

Thus the angle of prism P2P_2P2​ is

3∘3^\circ3∘

which corresponds to Option C.

  1. Comparison with stored answer

Stored correct answer: C

This matches our derived answer.

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