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With the help of a 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 *n* _{1} & *n* _{2} (*n* _{2} > *n* _{1}) respectively. Using this diagram, derive the relation

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

Write the sign conventions used. What happens to the focal length of the convex lens when it is immersed in water?

Dear Student ,

Here in this case let us draw the diagram first .

OM = u = object distance

MI = v =image distance

MC = R = radius of curvature

angle i = angle of incidence

angle r = angle of refraction

ON =incident ray

NI =refracted ray

NC = normal & n_{1} ,n_{2} are the refractive indices

From the figure with small angles

$\mathrm{tan}\angle NOM=\frac{MN}{OM},\mathrm{tan}\angle NCM=\frac{MN}{MC}and\mathrm{tan}\angle NIM=\frac{MN}{MI}\phantom{\rule{0ex}{0ex}}inthetriangleNOC,\angle i=exteriorangle=sumoftheinterioroppositeangle\phantom{\rule{0ex}{0ex}}\angle i=\angle NOC+\angle NCO=\angle NOM+\angle NOM=\frac{MN}{OM}+\frac{MN}{MC}\phantom{\rule{0ex}{0ex}}Similarly\angle NCM=\angle CNI=\angle CNI+\angle NIM\phantom{\rule{0ex}{0ex}}\angle r=\angle CNI=\angle MCN-\angle NIM=\frac{MN}{MC}-\frac{MN}{MI}\phantom{\rule{0ex}{0ex}}FromSnell\text{'}slaw,{n}_{1}\mathrm{sin}i={n}_{2}\mathrm{sin}r\phantom{\rule{0ex}{0ex}}Forsmallangles{n}_{1}i={n}_{2}r\phantom{\rule{0ex}{0ex}}Substitutingthevaluesofirweget,\phantom{\rule{0ex}{0ex}}{n}_{1}\left(\frac{MN}{OM}+\frac{MN}{MC}\right)={n}_{2}\left(\frac{MN}{MC}-\frac{MN}{MI}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{n}_{1}}{OM}+\frac{{n}_{2}}{MI}=\frac{{n}_{2}-{n}_{1}}{MC}\phantom{\rule{0ex}{0ex}}Applyingsigncnvention,OM=-u,MI=vandMC=R\phantom{\rule{0ex}{0ex}}\frac{{n}_{1}}{v}-\frac{{n}_{2}}{u}=\frac{{n}_{2}-{n}_{1}}{R}$

The **sign convention used here is as follows :**

(i) All the distances are measured from the optical centre that is pole .

(ii) If the distance is measured in the opposite direction of the incident rays then the distance is negative and in the same direction of incident rays the distances measured will be positive .

Here the object distance is negative and the image distance and the radius of curvature is positive .

Now when the convex lens is immersed in water then straight light ray pass from the glass to water and hence the refractive index of the water glass system is ,

$\frac{Velocityoflightinwater}{velocityoflightinglass}=\frac{2\xb7255\times {10}^{8}}{2\times {10}^{8}}=1\xb71275$

So more the refractive index more the light beam bend away from the normal while travelling from high density medium to low density medium and hence focal length is less .

So when the convex lens is immersed in water is less than that of lens in air so according to the refractive value of focal will shift and the focal will increase here when the convex lens is immersed in water .

Regards

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