For refraction at a curved surface- the u-v relationship is- -2-v-1-u-2-1-RA- -2 is the refractive index of the medium- at which the refraction takes place-B- -1 is the refractive index of the medium- from where the light is incident on the surface-C- R is the radius of curvature of the refractive surface-D- R can be positive or negative-

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Standard X

Physics

Refraction at Curved Surfaces

For refractio...

Question

For refraction at a curved surface, the u-v relationship is: $\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$

μ_{2}

is the refractive index of the medium, at which the refraction takes place.

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μ_{1}

is the refractive index of the medium, from where the light is incident on the surface.

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R is the radius of curvature of the refractive surface.

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R can be positive or negative.

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Solution

The correct options are
A $μ_{2}$ is the refractive index of the medium, at which the refraction takes place.
B $μ_{1}$ is the refractive index of the medium, from where the light is incident on the surface.
C R is the radius of curvature of the refractive surface.
D R can be positive or negative.

We see from the diagram that, $μ_{2}$ is the refractive index of the medium in which the refraction takes place and $μ_{1}$ is the refractive index of the medium, from where the light is incident on the surface. R is the radius of curvature of the surface and it can be positive or negative, given the surface is concave or convex.

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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. Where would an object be placed in a medium of refractive index

μ_{1}

, so that its real image is formed at equidistant from sphere (of radius R and refractive index

μ_{2}

) which is also placed in the medium of refractive index

μ_{1}

as shown in figure?

Q. A right angled prism of refractive index

μ_{1}

is placed in a rectangular block of refractive index

μ_{2}

, which is surrounded by a medium of refractive index

μ_{3},

as shown in the figure. A ray of light

^{'} e^{'}

enters the rectangular block at normal incidence. Depending upon the relationship between

μ_{1}, μ_{2}

and

μ_{3},

it takes one of the four possible paths

^{'} e f^{'},^{'} e g^{'},^{'} e h^{'}

^{'} e i^{'}

Match the paths in

List -I

with conditions of refractive indices in

List -II

and choose the correct answer.
List - I List - II

$(P) e \to f$	$(1) μ_{1} > \sqrt{2} μ_{2}$
$(Q) e \to g$	$(2) μ_{2} > μ_{1} & μ_{2} > μ_{3}$
$(R) e \to h$	$(3) μ_{1} = μ_{2} = μ_{3}$
$(S) e \to i$	$(4) μ_{2} < μ_{1} < \sqrt{2} μ_{2} & μ_{2} > μ_{1}$

A thin lens made of a material of refractive index $μ_{2}$ , has a medium of refractive index $μ_{1}$ on one side and a medium of refractive index $μ_{2}$ on the other side. The lens is biconvex and the two radii of curvature have equal magnitude R. A beam of light travelling parallel to the principal axis is incident on the lens. Where will the image be formed if the beam is incident from

(a) The medium $μ_{1}$ , and

(b) From the medium $μ_{3}$ ?

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For refraction at a curved surface, the u-v relationship is: μ2v−μ1u=μ2−μ1R

For refraction at a curved surface, the u-v relationship is: $\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$