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Worked Examples · Example 9.5

Q.Light from a point source in air falls on a spherical glass surface (n=1.5n = 1.5 and radius of curvature =20 cm= 20\ \text{cm}). The distance of the light source from the glass surface is 100 cm100\ \text{cm}. At what position the image is formed?

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Using the refraction formula for a single spherical surface, the image forms 100 cm100\ \text{cm} behind the glass surface, on the opposite side from the object — a real image.

The problem is a classic application of refraction at a single spherical surface. Light travels from air (refractive index n1=1n_1 = 1) into glass (n2=1.5n_2 = 1.5). The surface is convex toward the incident light (the source is in air, the glass is on the other side). The radius of curvature R=+20 cmR = +20\ \text{cm} by the sign convention: the centre of curvature lies on the side where light goes (into the glass), so RR is positive.

The key formula that governs this is the single spherical surface equation, which relates object distance uu, image distance vv, the two refractive indices, and the radius of curvature.

n2v−n1u=n2−n1R\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}

Here, n1=1n_1 = 1 (air), n2=1.5n_2 = 1.5 (glass), R=+20 cmR = +20\ \text{cm}, and u=−100 cmu = -100\ \text{cm} (object is real, on the incident side, so uu is negative by the Cartesian sign convention).

Let’s work through it step by step.

  1. Set up the equation with signs. The object is real and placed in front of the surface, so u=−100 cmu = -100\ \text{cm}. The surface is convex toward the object, so R=+20 cmR = +20\ \text{cm}. Substitute into the formula:

1.5v−1−100=1.5−120\frac{1.5}{v} - \frac{1}{-100} = \frac{1.5 - 1}{20}

  1. Simplify the right-hand side.

0.520=140\frac{0.5}{20} = \frac{1}{40}

  1. Simplify the left-hand side. The term −1−100-\frac{1}{-100} becomes +1100+\frac{1}{100}:

1.5v+1100=140\frac{1.5}{v} + \frac{1}{100} = \frac{1}{40}

  1. Isolate 1.5v\frac{1.5}{v}.

1.5v=140−1100\frac{1.5}{v} = \frac{1}{40} - \frac{1}{100}

Find a common denominator (200):

140=5200,1100=2200\frac{1}{40} = \frac{5}{200}, \quad \frac{1}{100} = \frac{2}{200}

So:

1.5v=5−2200=3200\frac{1.5}{v} = \frac{5 - 2}{200} = \frac{3}{200}

  1. Solve for vv.

1.5×2003=v1.5 \times \frac{200}{3} = v

v=3003=100 cmv = \frac{300}{3} = 100\ \text{cm} …

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