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An equilateral triangular prism has a refractive index of 1.5. Calculate the angle of minimum deviation for light passing through it.

Physics

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B. 37.2∘37.2^\circ

  1. Identify the given values.

    The prism is an equilateral triangular prism, so the prism angle A=60∘A = 60^\circ.
    The refractive index of the glass n=32=1.5n = \frac{3}{2} = 1.5.

  2. Use the formula for the refractive index of a prism at minimum deviation.

    The formula is:

    n=sin⁡(A+δm2)sin⁡(A2)n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}

    where δm\delta_m is the angle of minimum deviation.

  3. Substitute the known values into the formula.

    1.5=sin⁡(60∘+δm2)sin⁡(60∘2)1.5 = \frac{\sin\left(\frac{60^\circ + \delta_m}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)}

  4. Simplify the denominator.

    sin⁡(60∘2)=sin⁡(30∘)=0.5\sin\left(\frac{60^\circ}{2}\right) = \sin(30^\circ) = 0.5

  5. Substitute the simplified denominator back into the equation and solve for the sine term.

    1.5=sin⁡(60∘+δm2)0.51.5 = \frac{\sin\left(\frac{60^\circ + \delta_m}{2}\right)}{0.5} sin⁡(60∘+δm2)=1.5×0.5\sin\left(\frac{60^\circ + \delta_m}{2}\right) = 1.5 \times 0.5 sin⁡(60∘+δm2)=0.75\sin\left(\frac{60^\circ + \delta_m}{2}\right) = 0.75

  6. Find the angle by taking the inverse sine.

    60∘+δm2=arcsin⁡(0.75)\frac{60^\circ + \delta_m}{2} = \arcsin(0.75) 60∘+δm2≈48.59∘\frac{60^\circ + \delta_m}{2} \approx 48.59^\circ

  7. Solve for δm\delta_m.

    60∘+δm=2×48.59∘60^\circ + \delta_m = 2 \times 48.59^\circ 60∘+δm=97.18∘60^\circ + \delta_m = 97.18^\circ δm=97.18∘−60∘\delta_m = 97.18^\circ - 60^\circ δm=37.18∘\delta_m = 37.18^\circ

  8. Round the result to one decimal place and compare with the options.

    δm≈37.2∘\delta_m \approx 37.2^\circ.

    The closest option is B.

    The final answer is \boxed{\text{37.2^\circ}}.

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