Question 275: Ray Optics

Question

Light from a point source in air falls on a spherical glass surface (refractive index, $\mu=1.5$ and radius of curvature $=50 \mathrm{~cm}$ ). The image is formed at a distance of 200 cm from the glass surface inside the glass. The magnitude of distance of the light source from the glass surface is ___________m.

Accepted Answer

We apply the refraction formula at a spherical surface: ÃÂ¼Ã¢ÂÂ/v Ã¢ÂÂ ÃÂ¼Ã¢ÂÂ/u = (ÃÂ¼Ã¢ÂÂ Ã¢ÂÂ ÃÂ¼Ã¢ÂÂ)/R.
Here:
Ã¢ÂÂ¢ ÃÂ¼Ã¢ÂÂ = 1 (air)
Ã¢ÂÂ¢ ÃÂ¼Ã¢ÂÂ = 1.5 (glass)
Ã¢ÂÂ¢ R = 50 cm (radius of curvature)
Ã¢ÂÂ¢ v = +200 cm (image distance inside glass)
Ã¢ÂÂ¢ u = Ã¢ÂÂx (object distance, negative for a real object in front of the surface)
Substitute these values and solve for x:
$$\begin{aligned} & \frac{\mu_2}{\mathrm{v}}-\frac{\mu_1}{\mathrm{u}}=\frac{\mu_2-\mu_1}{\mathrm{R}} \ & \frac{1.5}{200}-\frac{1}{-\mathrm{x}}=\frac{1.5-1}{50} \ & \frac{1}{\mathrm{x}}=\frac{1}{100}-\frac{3}{400} \ & \mathrm{x}=400 \mathrm{~cm} \ & \mathrm{x}=4 \mathrm{~m} \end{aligned}$$
So, the light source is 4 m from the glass surface.

Ray Optics

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