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NEET PHYSICSMedium

A thin prism of angle 1515^{\circ} made of glass of refractive index μ1=1.5\mu_1=1.5 is combined with another prism of glass of refractive index μ2=1.75\mu_2=1.75. The combination of the prism produces dispersion without deviation. The angle of the second prism should be:

A

77^{\circ}

B

1010^{\circ}

C

1212^{\circ}

D

55^{\circ}

Step-by-Step Solution

  1. Condition for Dispersion without Deviation: For a combination of two thin prisms to produce dispersion without any net deviation, the mean deviation produced by the first prism must be equal and opposite to the mean deviation produced by the second prism.
  2. Formula: The deviation (δ\delta) produced by a thin prism of angle AA and refractive index μ\mu is given by δ=(μ1)A\delta = (\mu - 1)A.
  3. Calculation: For the first prism: δ1=(μ11)A1=(1.51)×15=0.5×15=7.5\delta_1 = (\mu_1 - 1)A_1 = (1.5 - 1) \times 15^{\circ} = 0.5 \times 15^{\circ} = 7.5^{\circ}. For the second prism: δ2=(μ21)A2=(1.751)A2=0.75A2\delta_2 = (\mu_2 - 1)A_2 = (1.75 - 1)A_2 = 0.75 A_2. Equating the magnitudes for zero net deviation: δ1=δ2\delta_1 = \delta_2. 7.5=0.75A27.5^{\circ} = 0.75 A_2.
  • A2=7.50.75=10A_2 = \frac{7.5}{0.75} = 10^{\circ}.
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