Calculate the angle of minimum deviation of a 60º prism of a refractive index 1.5[ $sin^{-1}$ (0.75) = 49º]

1. Analyze the question

We are given:

• The refracting angle of the prism ($A$) = $60^\circ$
• The refractive index of the prism ($n$) = $1.5$
• We need to find the angle of minimum deviation ($D_m$).

2. Explain the concept

For a prism, the relationship between the refractive index ($n$), the refracting angle ($A$), and the angle of minimum deviation ($D_m$) is given by the formula:

3. Show the solution

Step 1: Substitute the known values into the formula.

Step 2: Simplify the denominator.

Step 3: Solve for the sine term.

Step 4: Use the inverse sine to find the angle. The question provides the hint: $\sin^{-1}(0.75) = 49^\circ$.

Step 5: Solve for $D_m$.

4. Identify the answer

Comparing our result ($38^\circ$) to the given options, it matches option A.

5. Tip

When dealing with prism problems, always remember that at minimum deviation, the light ray passes symmetrically through the prism. This symmetry is why the angle of incidence equals the angle of emergence, which simplifies the math significantly!