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(i) Draw a ray diagram showing the geometery of formation of image of a point object situated on the principal axis and on the convex side of a spherical surface of radius of curvature R. Taking the rays as incident from a rarer medium of refractive index n₁ to a denser medium of refractive index n₂ , derive the relation.

$\frac{n_{2}}{v}-\frac{n_{1}}{u}=\frac{n_{2}-n_{1}}{R}$ , where symbols have their usual meaning.

(ii) Explain briefly how the focal length of a convex lens changes with increase in wavelength of incident light.

(iii) What happens to the focal length of convex lens when it is immersed in water? Refractive index of the material of lens is greater than that of water.

Answers (1)

(i)

from $\Delta OMN$

$\tan \angle NOM = \frac{MN}{-u}$ .....(1)

from $\Delta MNC$

$\tan \angle NCM = \frac{MN}{R}$ .......(2)

form $\Delta MNI$

$\tan \angle NIM = \frac{MN}{V}$ .......(3)

The angle is very small

(1) + (2)

$\because i< NOM + NCM = \frac{-MN}{-u}+\frac{MN}{R}$ .......(4)

$\angle NCM = \angle NIM +r$

$r=\angle NCM-\angle NIM$

$r=\frac{MN}{R}-\frac{MN}{V}$ .....(5)

from Snell's law

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

$n_{1} i= n_{2} r$

$n_{1} \left ( \frac{MN}{-u} +\frac{MN}{R}\right )= n_{2} \left ( \frac{MN}{R}-\frac{MN}{V} \right )$

or

$\frac{n_{2}-n_{1}}{R}=\frac{n_{2}}{V}-\frac{n_{1}}{a}$

(ii) $\frac{1}{f}=\left ( \frac{\mu _{2}}{\mu _{1}}-1 \right )\frac{2}{R}as \frac{\mu _{2}}{\mu _{1}}$ decreases, focal length increases

(iii) as $\mu _{1}$ increases focal length increases.

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Safeer PP

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