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Draw a ray diagram showing the geometry 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 $n_1$ to a denser medium of refractive index $n_2$ derive the relation. $\large\frac{n_@}{v} -\frac{n_1}{u} = \frac{n_2-n_1}{R}$, Where symbols have their usual meaning ? (ii) Explain briefly how the focal length of a convex lens changes with increase in wave length 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 water .

1 Answer

Solution :
When the object O in the rarer medium lies close to the pole P of the convex refracting surface, the two refractive rays appear to diverge from a point I on the principal again So I is the virtual image of the point object O.
In $\Delta NOC, i=2+r$
In $\Delta NCI , r = \beta + r$
If all the rays are par axial then all the angle will be small , then
$\alpha = \tan \alpha = \large\frac{NM}{IM} = \frac{NM}{OP}$
$\beta = \tan \beta = \large\frac{NM}{IM} = \frac{NM}{IP}$
$r = \tan \gamma = \large\frac{NM}{MC} = \frac{NM}{PC}$
From Snell's law $ \large\frac{\sin i }{\sin r} = \frac{n_2}{n_1}$
But since i and r are small, $\large\frac{i}{r} =\frac{n_2}{n_1}$
or $n_1( \alpha + r)= n_2(\beta+r)$
$n_1 \bigg[ \large\frac{NM}{OP}+\frac{NM}{PC}\bigg]$$ = n_2 \bigg[ \large\frac{NM}{IP} +\frac{NM}{PC}\bigg]$
$n_1 \bigg[ \large\frac{1}{OP}+\frac{1}{PC}\bigg]$$ = n_2 \bigg[ \large\frac{1}{IP} +\frac{1}{PC}\bigg]$
=> $ \large\frac{n_1}{OP} -\frac{n_2}{IP} = \frac{n_2 - n_1}{PC} $
=> But $OP=-u, IP= -v$ an d$PC= +R$
$\therefore \frac{n_1}{-u} - \frac{n_2}{-v} =\frac{n_2-n_1}{R}$
$\therefore \frac{n_2}{v} - \frac{n_1}{u} =\frac{n_2-n_1}{R}$
(ii)
Since $f= \large\frac{1}{\mu -1}$
For $ \lambda_2 > \lambda_1 , \mu_2 < \mu_1$
Hence $f_2 > f_1$
Therefore When wave length of incident light is increased , focal length of the convex lens increases.
(iii)
According to lens maker's formula
$\large\frac{1}{f} $$ = (\mu-1) \bigg( \large\frac{1}{R_1}-\frac{1}{R_2}\bigg)$
(ie) $f \alpha \large\frac{1}{\mu-1}$
As $^{w} \mu g < ^a \mu g$, the focal length of the lens will increase when it is completely immersed in water .
answered Jun 10, 2016 by meena.p
 

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