2019 A/L Physics Past Paper · Question 43 (English Medium)

Three right-angled prisms: the incorrect expression

Three right angled prisms of refractive indices $n_1, n_2, n_3\ (n_2 > n_1, n_3)$ are arranged very close to each other on a table as shown in the figure. There are no gaps between the contact surfaces of the prisms. A ray entering through the face $AB$ with an incident angle $i$ , refracts at faces $AB, BC, CD, DE$ , and emerges from the face $DE$ without deviation. The angles of refraction at the faces $AB, BC, CD$ are $r_1, r_2, r_3$ , respectively. Which of the following expressions is incorrect?

(1) $\sin i = n_1\sin r_1$
(2) $n_2\sin r_2 = n_1\cos r_1$
(3) $\sin i = n_3\cos r_3$
(4) $n_2\cos r_2 = n_3\sin r_3$
(5) $\cos i = n_3\cos r_3$

Answer and explanation

Answer: (3) $\sin i = n_3\cos r_3$

What the question asks

We want:

Which of the five expressions is not correct.

We follow the ray through the four faces $AB$ , $BC$ , $CD$ and $DE$ .

Relevant theory

(2) For all rays of light of a given colour passing through two given media, the sine of the angle of incidence bears a constant ratio to the sine of the angle of refraction.

(1) The incident ray, the refracted ray and the normal to the interface at the point of incidence all lie on the same plane.

The corners at $B$ , $C$ and $D$ are right angles.

So $AB \perp BC$ , $BC \perp CD$ and $CD \perp DE$ .

That makes $AB \parallel CD$ and $BC \parallel DE$ .

If the ray bends to the angle $r$ at one face, the next face is at a right angle to it.

So the angle of incidence there is $90° - r$ .

Reading the options

The five expressions are Snell’s law at the four faces, or two of them put together.

We check each one.

Reasoning and calculation

At every face:

$n_{before}\sin(\text{angle of incidence}) = n_{after}\sin(\text{angle of refraction})$

Faces next to each other are at a right angle.

$\sin i = n_1\sin r_1$

$n_1\sin(90^\circ - r_1) = n_2\sin r_2 \Rightarrow n_1\cos r_1 = n_2\sin r_2$

$n_2\cos r_2 = n_3\sin r_3$

$n_3\cos r_3 = \sin e$

$n_3\cos r_3 = \cos i$

Compare:

Expressions 1, 2, 4 and 5 agree with the equations above.

Expression 3 is $\sin i = n_3\cos r_3$ .

It takes $e = i$ .

But $DE$ is at a right angle to $AB$ , not parallel to it.

So only expression 3 is left.

If $e = i$ the ray would leave tilted.

Without deviation we need $e = 90° - i$ .

So only 3 is not correct.

Why the other options fail

Snell’s law at $AB$ . It is correct.

Snell’s law at $BC$ . The angle of incidence is $90° - r_1$ . It is correct.

Snell’s law at $CD$ . The angle of incidence is $90° - r_2$ . It is correct.

The ray leaves $DE$ without deviation, so $e = 90° - i$ . It is correct.

Conclusion

Check a special case:

Let $i = 0$ . The ray enters straight and $r_1 = 0$ .

At $BC$ the angle of incidence is $90°$ . The ray runs along $BC$ .

Expression 5 then gives $n_3\cos r_3 = 1$ , so $e = 90°$ .

That is the same direction as the entering ray ✓.

Expression 3 would give $n_3\cos r_3 = 0$ . That cannot be.

$DE$ is at a right angle to $AB$ . So without deviation $\sin e = \cos i$ .

That gives $n_3\cos r_3 = \cos i$ , which is correct.

$\sin i = n_3\cos r_3$ is not correct.

So the answer is option (3).

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