A thin beam of light enters a thick plastic sheet from air at an angle of 32.0° with the normal and continues in the sheet at an angle of 23.0° with the normal. What is the index of refraction of the plastic? The index of refraction of the air, nair = 1.

Question

A thin beam of light enters a thick plastic sheet from air at an angle of 32.0° with the normal and continues in the sheet at an angle of 23.0° with the normal.  What is the index of refraction of the plastic? The index of refraction of the air, nair = 1.

BlackTom AI · Accepted Answer

We start by identifying the setup and applying Snell's law at the air–plastic interface. Snell's law relates the indices of refraction and the angles of the incident and refracted rays: n1 sin(theta1) = n2 sin(theta2).

Given: n1 (air) = 1.0, theta1 = 32.0° (angle with the normal in air), theta2 = 23.0° (angle with the normal inside the plastic).

Solve for n2 (the plastic's index of refraction): n2 = n1 * sin(theta1) / sin(theta2).

Compute the sines: sin(32.0°) ≈ 0.5299, sin(23.0°) ≈ 0.3907.

Plug in the values: n2 ≈ 1.0 * 0.5299 / 0.3907 ≈ 1.357.

Rounding to three significant figures gives n2 ≈ 1.36.

Notes:
- The calculation assumes the light indeed refracts into the plastic, so the refracted angle is smaller than the incident angle, consistent with a medium of higher index than air.
- If answer options were provided, you would compare this computed value to the closest choice; since options aren’t listed here, the numerical result is the key outcome.
- The method relies on the standard form of Snell's law and uses the given air index nair = 1 as the reference medium.

PHYSICS 1201 SU2025 (10585)

A thin beam of light enters a thick plastic sheet from air at an angle of 32.0° with the normal and continues in the sheet at an angle of 23.0° with the normal. What is the index of refraction of the plastic? The index of refraction of the air, nair = 1.

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