📘 Prism & Dispersion of Light – Worksheet (20 Questions)

🔷 Section A: Theory & Conceptual (10 Questions)

Q1.

Define angle of deviation in a prism. On what factors does it depend?

Q2. (Diagram-Based)

A ray passes through a prism:


        /\
       /  \
 i →  /    \ → e
     /______\

Mark the following on the diagram:

Angle of incidence (i)
Angle of emergence (e)
Angle of prism (A)
Angle of deviation (δ)

Q3.

State the condition for minimum deviation in a prism.

Q4.

Why does a prism disperse white light?

Q5.

Which colour deviates the most and why?

Q6.

Explain why red light deviates least in a prism.

Q7.

Define dispersive power of a prism.

Q8.

What is meant by angular dispersion?

Q9. (Diagram-Based)

White light passes through a prism and forms a spectrum:


White light → /\
             /  \
            /____\ → VIBGYOR

Explain why the spectrum spreads out.

Q10.

Is deviation possible without dispersion? Give example.

🔶 Section B: Numerical Problems (10 Questions)

Q11.

A prism has refracting angle $A = 60^\circ$ . The angle of minimum deviation is $40^\circ$ .
Find the refractive index.

Q12.

For a prism,

$\mu = 1.5,\quad A = 60^\circ$

Find the angle of minimum deviation.

Q13.

The refractive indices for violet and red light are 1.54 and 1.50 respectively.
Find the dispersive power of the prism.

Q14.

Angle of prism = $60^\circ$ , angle of incidence = $50^\circ$ , angle of emergence = $60^\circ$ .
Find the angle of deviation.

Q15.

A prism produces an angular dispersion of $2^\circ$ .
If mean deviation is $40^\circ$ , find dispersive power.

Q16.

Refractive index varies with wavelength. If

$\mu_v > \mu_r$

Explain quantitatively which deviates more.

Q17.

For a prism at minimum deviation, the angle of refraction inside the prism is $30^\circ$ .
Find the angle of prism.

Q18.

If the refractive index of a prism is 1.73 and angle of prism is $60^\circ$ ,
find the angle of minimum deviation.

Q19.

A ray suffers deviation of $30^\circ$ in a prism of angle $50^\circ$ .
Find the sum of angles of incidence and emergence.

Q20. (Diagram-Based Numerical)


        /\
       /  \
      /____\

Given:

$A = 60^\circ$
$i = 45^{\circ}$
$r_1 = 30^\circ$

Find:

$r_2$
Angle of emergence

✅ Answer Key

Section A:

Angle between incident and emergent ray
Diagram labeling
$i = e$ , symmetric path
Different refractive index for different wavelengths
Violet (higher μ)
Lower refractive index
Ratio of angular dispersion to mean deviation
Difference in deviation of colours
Different μ for different wavelengths
Yes (e.g., monochromatic light)

Section B:

$\mu = 1.53$
$\delta_m \approx 37^\circ$
$\omega = 0.027$
$\delta = 50^\circ$
$ω = 0.05$
Violet deviates more
$A = 60^\circ$
$\delta_m = 60^\circ$
$i + e = 80^{\circ}$
$r_2 = 30^\circ$ , solve for $e$

🧠 Detailed Solutions

Q11 Solution

Use formula:

$\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$

$\mu = \frac{\sin(50^\circ)}{\sin(30^\circ)} = \frac{0.766}{0.5} = 1.53$

Q12 Solution

$\delta_m = 2 \sin^{-1}(\mu \sin A/2) - A$ $= 2\sin^{-1}(1.5 \times 0.5) - 60$ $= 2 \times 48.6 - 60 \approx 37^\circ$

Q13 Solution

$\omega = \frac{\mu_v - \mu_r}{\mu_y - 1}$ $= \frac{1.54 - 1.50}{1.52 - 1} = 0.027$

Q14 Solution

$\delta = i + e - A = 50 + 60 - 60 = 50^\circ$

Q15 Solution

$\omega = \frac{\text{angular dispersion}}{\text{mean deviation}} = \frac{2}{40} = 0.05$

Q16 Solution

Since $\mu_v > \mu_r$ , deviation:

$\delta_v > \delta_r$

Q17 Solution

At minimum deviation:

$r_1 = r_2 = \frac{A}{2} \Rightarrow A = 60^\circ$

Q18 Solution

$\delta_m = 2\sin^{-1}(1.73 \times 0.5) - 60 = 2\sin^{-1}(0.866) - 60 = 2 \times 60 - 60 = 60^\circ$

Q19 Solution

$\delta = i + e - A \Rightarrow i + e = \delta + A = 30 + 50 = 80^\circ$

Q20 Solution

$r_1 + r_2 = A \Rightarrow r_2 = 30^\circ$

Using Snell’s law:

$\mu = \frac{\sin i}{\sin r}$

Then find angle of emergence.

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