A 10 cm thick flat slab (n = 1.49) is placed in the air. A ray of light hits the front surface of the flat slab with an incident angle of 60 degrees. How much is the emergent ray displaced (d)?

To find the displacement \( d \) of the emergent ray as it exits the flat slab, we can use geometric optics principles. Here's how to approach the problem:

### Step 1: Refraction at the first surface

When a light ray passes from air (with refractive index \( n_1 = 1 \)) into a medium with refractive index \( n_2 = 1.49 \), it bends according to Snell's law:

\[
n_1 \sin \theta_1 = n_2 \sin \theta_2
\]

where:
- \( n_1 = 1 \) is the refractive index of air,
- \( n_2 = 1.49 \) is the refractive index of the slab,
- \( \theta_1 = 60^\circ \) is the angle of incidence,
- \( \theta_2 \) is the angle of refraction in the slab.

Let's solve for \( \theta_2 \):

\[
\sin \theta_2 = \frac{n_1 \sin \theta_1}{n_2} = \frac{1 \times \sin 60^\circ}{1.49}
\]

Since \( \sin 60^\circ = \frac{\sqrt{3}}{2} \):

\[
\sin \theta_2 = \frac{\frac{\sqrt{3}}{2}}{1.49}
\]

Now calculate \( \theta_2 \).

This problem explores the displacement of a light ray passing through a 10 cm thick slab with a refractive index of 1.49. Using Snell's law and geometric optics, the solution calculates the angle of refraction and the lateral displacement of the ray. Suitable for high school students in Grades 11-12.

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