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# Refraction of light at a spherical surface Contents...

## 1.For real image :-

The figure shows the geometry of formation of image I of an object O on the principal axis of a spherical surface ( convex ) with centre of curvature C and radius of curvature R.

The rays are incident from a medium of refractive index μ₁ to another of refractive index μ₂.

Suppose that, ∠SCP = α, ∠SIP = β and ∠SOP = θ

Now, from ∆SOC, i = α +θ and from ∆ SCI, α = β + r

or r = α – β

Now, from Snell’s law

μ₁ sin i = μ₂ sin r

for paraxial rays-

μ₁ i = μ₂r

or μ₁ ( α + β) = μ₂ ( α – β )

or μ₁ ( SP/PC + SP/PO ) = μ₂ ( SP/PC – SP/PI )

or μ₁ ( 1/R + 1/-u ) = μ₂ ( 1/R – 1/v )

since, according to the sign convention u is negative, v and R are positive.

or μ₁/R – μ₁/u = μ₂/R – μ₂/v

Hence, μ₂/v – μ₁/u = ( μ₂ – μ₁ )/R

μ₁ = 1 ( Air ) and μ₂ = μ (Glass ), then

μ/v – 1/u = ( μ – 1 )/R

## 2. For virtual image :-

Let us consider, refractive index of medium 1 = μ₁

Refractive index of medium 2 = μ₂

Object distance PO = – u

Image distance PI = – v

Radius of curvature PC = +R

Angle of incident = i, angle of refraction = r

From the above figure,

In ∆ SOC, i = α + θ and in ∆SIC, r = α + β,

Now from Snell’s law

μ₁ sin i = μ₂ sin r

For paraxial rays, μ₁ i = μ₂ r

or μ₁ ( α + θ ) = μ₂ ( α + β )

or μ₁ ( SP/PC + SP/PO ) =μ₂ ( SP/PC + SP/PI )

μ₁ ( 1/R + 1/-u ) = μ₂ ( 1/R + 1/-v )

or μ₁/R – μ₁/u = μ₂/R – μ₂/v

Hence, μ₂/v – μ₁/u = ( μ₂ – μ₁ )/R

If μ₁ = 1 ( Air ) and μ₂ = μ ( Glass ), then

μ/v – 1/u = ( μ – 1 )/R

## Refraction of light at a concave surface :-

Let us consider, refractive index of medium 1 = μ₁

Refractive index of medium 2 = μ₂

Object distance PO = – u

Image distance PI = – v

Radius of curvature PC = – R

Angle of incident = i

Angle of refraction = r

From the above figure,

∠SCP = α, ∠SIP = β and ∠SOP = θ

Now, from ∆SOC, α = i + θ or i = α – θ and

from ∆SIC, α = r + β or r = α – β

Now, from Snell’s law

μ₁ sin i = μ₂ sin r

For paraxial rays, μ₁ i = μ₂ r

or μ₁ ( α -θ ) = μ₂ ( α – β )

or μ₁ ( SP/PC – SP/PO ) = μ₂ ( SP/PC – SP/PI )

or μ₁ ( 1/- R – 1/- u ) = μ₂ ( 1/- R – 1/- v )

or ₋ μ₁/R + μ₁/u = -μ₂/R + μ₂/v

Hence, μ₂/v – μ₁/u = ( μ₂ – μ₁ )/R

Numerical question-

1. Light from a point source in air falls on a spherical glass surface ( μ = 1·5 and radius of curvature =20 cm ). The distance of the light source from the glass surface is 100 cm. At what position the image is formed ?

Solution.

μ₁ = 1 ( refractive index of air = 1 )

μ₂ = 1·5, R = 20 cm, u = – 100 cm

v = ?

μ₂/v – μ₁/u = ( μ₂ – μ₁ )/R

1·5/v + 1/100 = ( 1·5 – 1 )/20

1·5/v = 5/200 – 1/100 = ( 5 – 2 )/200 = 3/200

v = ( 1·5 x 200 )/3 = + 100 cm

Image formed at a distance of 100 cm in the direction of incident light.

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