# Refraction from a semi-circular disc

Refraction is bending of light as it passes from one medium to another medium. A semicircular disc can be used to verify laws of refraction:

1. The incident ray, refracted ray, and normal lies in the same plane.
2. The angle of incidence $$(i)$$, angle of refraction $$(r)$$ and the refractive index of the medium ($$\mu$$) are related by $${\sin i}/{\sin r}=\mu$$.

In this experiment, you need a semi-circular disc, laser light, white paper, protector, pencil, and a scale.

1. Place the semi-circular disc on the paper. Draw its boundary. Find centre of the face and draw normal to it.
2. Draw three lines making an angle of 30 degree, 45 degree, and 60 degree with the normal.
3. Place the laser such that incident ray is parallel to one of the line and it strike the semi-circular disc at centre of rectangular face. Observe the refracted light. Mark two points on the path of refracted line and join them. Measure angle of incidence, angle of refraction, and calculate $$\mu$$ using formula given above.
4. Repeat above step for other two angles.
5. Why incident ray should strike at the centre of rectangular face of semi-circular disc?

## Focal length of a semi-circular disc

Question: What is the focal length of a semi-circular disc of radius of curvature $R$ made from glass material of refractive index 1.5?

Answer: Substiting $R_1=R$, $R_2=\infty$ and $\mu=1.5$ in lens maker's formula \begin{align} \frac{1}{f}=(\mu-1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right] \end{align} will give $f=2R$. However, the lens maker's formula gives correct result only for thin lenses. Semi-circular disc is a thick lens. We experimentaly found that parallel rays converges at the perimeter of the disc. Thus, focal length of the semi-circular disc is $R$.

Question: What is the focal length of a sphere of radius $R$ made from a glass material of refractive index 1.5?

Answer: It is a thick lens. The lens maker's formula cannot be used. The focal length of the sphere of radius $R$ made from a material of refractive index 1.5 is $f=3R/2$. See solution (pdf).

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