Derive the relation between object distance (u) and image distance (v) in terms of refractive index of the medium (n) and the radius of curvature (R) of the spherical surface.

\frac{μ_{1}}{μ_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{μ_{2}}{μ_{1}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{μ_{2}}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{μ_{1}}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{μ_{2}}{μ_{1}}$
$α + β = i$
$r + r = β$
$\frac{α + β}{β = r} = \frac{i}{r}$
$⟹ \frac{tan α + tan β}{tan β - tan r} = \frac{sin i}{sin r} = \frac{μ_{2}}{μ_{1}}$
$\frac{\frac{P Q}{P u} + \frac{P Q}{P C}}{\frac{P Q}{P C} - \frac{P Q}{P V}} = \frac{μ_{2}}{μ_{1}}$

Suggest Corrections

Similar questions

Q. For refraction at a curved surface; the u.v relationship is:

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

Q. For refraction at a curved surface, the u-v relationship is:

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

Q. A plano convex lens of refractive index

μ_{1}

and focal length

f_{1}

is kept in contact with another plano concave lens of refractive index

μ_{2}

and focal length

f_{2}

. If the radius of curvature of their spherical faces is

R

each and

f_{1} = 2 f_{2}

, then

μ_{1}

and

μ_{2}

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

A plane-convex lens fits exactly into a plane-concave lens. Their plane surfaces are parallel to each other. If lenses are made of different materials of refractive index $μ_{1}$ and $μ_{2}$ and R is the radius of curvature of the curved surface of the lenses, then the focal length of the combination is

Join BYJU'S Learning Program

Derive the relation between object distance (u) and image distance (v) in terms of refractive index of the medium (n) and the radius of curvature (R) of the spherical surface.

The correct option is B μ2μ1α+β=ir+r=βα+ββ=r=ir⟹tanα+tanβtanβ−tanr=sinisinr=μ2μ1PQPu+PQPCPQPC−PQPV=μ2μ1