Question

With the help of ray diagram, show the formation of image of a point object by refraction of light at a spherical surface separating two media of refractive indices n₁and n₂ (n₂ > n₁) respectively. Using this diagram, derive the relation.

$\frac{n_{2}}{v} - \frac{n_{1}}{u} = \frac{n_{2} - n_{1}}{R}$

What happens to the focal length of convex lens when it is immersed in water?

Solution

(i) In the ray diagram below, AMB is a convex surface separating two media of refractive indices n₁ and n₂ (n₂> n₁). Consider a point object O placed on the principal axis. A ray ON is incident at N and refracts along NI. The ray along ON goes straight and meets the previous ray at I. Thus, I is the real image of O.

From Snell's law,

n_{2} = \frac{\sin i}{\sin r}

n_{1} \sin i = n_{2} \sin r \frac{n_{2}}{n_{1}} = \frac{\sin i}{\sin r}

\Rightarrow

n_{1} i = n_{2} r

∆ NOC,

i = α + γ

∆ NIC,

γ = r - β

r = γ - β

∴

n_{1} (α + γ) = n_{2} (γ - β)

i.e.,

n_{1} α + n_{2} β = (n_{2} - n_{1}) γ

But,

α ≅ \tan α = \frac{NP}{OP} = \frac{NP}{OM}

[P is close to M]

β ≅ \tan β = \frac{NP}{PI} = \frac{NP}{MI}

γ ≅ \tan γ = \frac{NP}{PC} = \frac{NP}{MC}

∴

n_{1} . \frac{NP}{OM} + n_{2} . \frac{NP}{MI} = (n_{2} - n_{1}) \frac{NP}{MC}

\Rightarrow

\frac{n_{1}}{OM} + \frac{n_{2}}{MI} = \frac{n_{2} - n_{1}}{MC}

Now, using Cartesian sign convention,

OM = - u, MI = + v, MC = + R

∴

\frac{n_{1}}{- u} + \frac{n_{2}}{v} = \frac{n_{2} - n_{1}}{R}

\Rightarrow

\frac{n_{2}}{v} - \frac{n_{1}}{u} = \frac{n_{2} - n_{1}}{R}

The lens makers formula gives us the relationship,

\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})

That is, focal length and refractive index has inverse dependence.
Refractive index of water is greater than that of air and so, focal length of the lens will reduce when immersed in water.

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