2021 A/L Physics Past Paper · Question 41 (English Medium)

Angles of refraction of red and blue rays at minimum deviation

Two light rays of red and blue colour are made to pass separately through a glass prism of refracting angle $60^\circ$ . If both rays suffer minimum deviation while passing the prism, which of the following is correct with regard to the angle of refraction of the red ray ( $r_R$ ) and the angle of refraction of the blue ray ( $r_B$ ) at the surface of incidence of the prism?

(1) $r_R > r_B$
(2) $r_R < r_B$
(3) $r_R = r_B \neq 30^\circ$
(4) $r_R = r_B = 30^\circ$
(5) $r_R = r_B = 60^\circ$

Answer and explanation

Answer: (4) $r_R = r_B = 30^\circ$

What the question asks

We observe: red and blue rays each pass through a prism of refracting angle $A = 60^\circ$ , each at its own minimum deviation.

Find how $r_R$ and $r_B$ compare at the first surface.

Relevant theory

r₁ + r₂ = A

At the state of minimum deviation the ray is found to pass symmetrically through the prism.

Required equations

$r_1 + r_2 = A$

$r_1 = r_2 = r$

$r = \dfrac{A}{2}$

Reading the options

The options compare $r_R$ with $r_B$ , and give a value.

Candidates: red larger, blue larger, equal at some value, equal at $30^\circ$ , equal at $60^\circ$ .

Reasoning and calculation

The constraints. At minimum deviation the passage is symmetric, so $r_1 = r_2$ . And $r_1 + r_2 = A$ .

Neither fact mentions the colour. So $r = \dfrac{A}{2}$ is the same for red and for blue.

What the colour changes is the refractive index $n$ . For blue $n$ is larger, so the angle of incidence $i$ and the deviation $D$ are larger. But $r$ is not.

$60$

$2r = 60^\circ$

$30$

$r_R = r_B = \dfrac{60^\circ}{2}$

$r_R > r_B$ confuses $r$ with $i$ . The colour changes $i$ , not $r$ .

$r_R < r_B$ confuses $r$ with $i$ . The colour changes $i$ , not $r$ .

Equal is right, but $r = \dfrac{A}{2}$ is fixed at $30^\circ$ .

$60^\circ$ is the whole angle $A$ . Each $r$ is half of it.

Checks

$a / b$

Do the limits make sense?

The prism angle $A$ and the symmetry fix $r$ . No colour appears in $r = \dfrac{A}{2}$ .

If the prism angle were $40^\circ$ , then $r$ would be $20^\circ$ for every colour. Only $i$ and $D$ would change with colour.

Conclusion

$r_R = r_B = \dfrac{A}{2} = 30^\circ$ .

The answer is (4).

At minimum deviation the path through the prism is symmetric. So $r = A/2$ , whatever the colour.

The colour changes the angle of incidence and the deviation. It does not change $r$ .

Exam tip

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