Question 293: Ray Optics

Question

Two transparent media having refractive indices 1.0 and 1.5 are separated by a spherical refracting surface of radius of curvature $$30 \mathrm{~cm}$$. The centre of curvature of surface is towards denser medium and a point object is placed on the principle axis in rarer medium at a distance of $$15 \mathrm{~cm}$$ from the pole of the surface. The distance of image from the pole of the surface is ____________ $$\mathrm{cm}$$.

Accepted Answer

The refraction at a spherical surface is governed by the formula:
$\frac{1}{v} - \frac{1}{u} = \frac{n_2 - n_1}{R} n_2$
where:
(v) is the image distance,
(u) is the object distance,
($n_1$) is the refractive index of the medium where the object is,
($n_2$) is the refractive index of the medium where the image is formed,
(R) is the radius of curvature of the refracting surface.
Here, we have:
(u = -15) cm (the object is on the same side as the light is coming from, so the distance is taken as negative),
($n_1$ = 1.0) (refractive index of the rarer medium),
($n_2$ = 1.5) (refractive index of the denser medium),
(R = -30) cm (the center of curvature is in the denser medium, so the radius is taken as negative).
Substituting these values into the formula, we get:
$\frac{1}{v} - \left(-\frac{1}{15}ight) = \frac{1.5 - 1.0}{-30} \cdot 1.5$
Solving for (v), we get:
$v = \frac{1}{\frac{1}{15} + \frac{1.5}{30}} = -30 \, 	ext{cm}$
Therefore, the image is formed at a distance of 30 cm from the pole of the surface, on the same side as the object. The negative sign indicates that the image is virtual and is formed on the same side of the surface as the light is coming from.

Ray Optics

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