Two transparent media of refractive indi...

Two transparent media of refractive indices `n_(1) and n_(2)` are separated by a spherical transparent surface. The rays of light incident on the surface get refracted into the medium on the other side. The laws of refraction are valid at each point of the spherical surface. A lens is a transparent optical medium bounded by two surfaces, at least one of which should be spherical. The focal length of a lens is determined by curvature `(R_(1) and R_(2))` of its two surfaces and the refractive index (n) of the medium of the lens with respect to the surrounding medium. Depending on `R_(1) and R_(2)`, a lens behaves as a diverging or a converging lens. The ability of a lens to diverge or converge a beam of light incident on it defines its power.
An object is placed at the point B as shown in the figure. The object distance (u) and the image distance (v) are related as

`1/v-1/u=((n_(2)-n_(1))/n_(1))1/R`

`1/v-1/u=((n_(1)-n_(2))/n_(2))1/R`

`n_(2)/v-n_(1)/u=((n_(2)-n_(1)))/R`

`n_(1)/v-n_(2)/u=((n_(1)-n_(2)))/R`

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Two refracting media are separated by a spherical interface as shown in the figure.

If `mu_(2) gt mu_(1)`, then there cannot be real image of real object.

If `mu_(2) gt mu_(1)`, then there cannot be real image of virtual object.

If `mu_(1) gt mu_(2)`, then there cannot be a virtual image of virtual object.

If `mu_(1) gt mu_(2)`, then there cannot be a real image of real object.

`58^(@)`

`29^(@)`

`32^(@)`

`42^(@)`

If `mu_2gtmu_1,` then there cannot be a real image of real object

If `mu_2gtmu_1,` then there cannot be a real image of virtual object

If `mu_1gtmu_2,` then there cannot be a real image of virtual object

If `mu_1gtmu_2,` then there cannot be a virtual image of virtual object