Question

A thin prism with angle 5° of refractive index 1.72 is combined with another prism of refractive index 1.9 to produce dispersion without deviation. The angle of second prism is :

• A. 4.5°
• B. 5°
• C. 4°
• D. 6°

Hint:

"Dispersion without deviation" requires the deviations to cancel out. "Deviation without dispersion" requires the angular dispersions (\(\delta_v - \delta_r\)) to cancel out.

Answer 1

The Correct Option is C

To find the angle of the second prism that allows dispersion without deviation, we need to understand the concept of combining prisms to produce dispersion without deviation. This means that the net angular deviation is zero, but there is a dispersion of light.

Here's the step-by-step solution:

1. For dispersion without deviation, the angular deviation caused by each prism should cancel each other out.
2. Let's denote:
   • \(A_1\) = angle of the first prism = 5°
   • \(\mu_1\) = refractive index of the first prism = 1.72
   • \(A_2\) = angle of the second prism = ?
   • \(\mu_2\) = refractive index of the second prism = 1.9
3. The deviation produced by a prism is given by the formula: \(\delta = (\mu - 1) \times A\)
4. For no net deviation: \((\mu_1 - 1) \times A_1 = (\mu_2 - 1) \times A_2\)
5. Substitute the known values into the equation: \((1.72 - 1) \times 5° = (1.9 - 1) \times A_2\)
6. Calculate the left side: \(0.72 \times 5° = 3.6°\)
7. Now, solve for \(A_2\): \(3.6° = 0.9 \times A_2\)
8. Therefore, \(A_2 = \frac{3.6°}{0.9} = 4°\)

Thus, the angle of the second prism should be 4° to achieve dispersion without deviation.

Therefore, the correct option is 4°.