Explain the refraction by two transparent spherically curved surfaces.
(N/A) spherical surface can be treated as a collection of infinitesimal planar elements. The laws of refraction (Snell's Law) apply at each point on the surface.
The normal at the point of incidence is perpendicular to the tangent plane at that point and always passes through the centre of curvature of the spherical surface.
When light passes through two such surfaces (as in a lens),the image formed by the first surface acts as the object for the second surface. By applying the refraction formula $n_2/v - n_1/u = (n_2 - n_1)/R$ sequentially to both surfaces,we derive the Lens Maker's Formula.
The normal at the point of incidence is perpendicular to the tangent plane at that point and always passes through the centre of curvature of the spherical surface.
When light passes through two such surfaces (as in a lens),the image formed by the first surface acts as the object for the second surface. By applying the refraction formula $n_2/v - n_1/u = (n_2 - n_1)/R$ sequentially to both surfaces,we derive the Lens Maker's Formula.
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Similar Questions
$A$ curved surface of radius $R$ separates two media of refractive indices $\mu_1$ and $\mu_2$ as shown in figures $A$ and $B$. Choose the correct statement$(s)$ related to the virtual image formed by object $O$ placed at a distance $x$,as shown in figure $A$.
Region $I$ and $II$ are separated by a spherical surface of radius $25\, \text{cm}$. An object is kept in region $I$ at a distance of $40\, \text{cm}$ from the surface. The distance of the image from the surface is $.....\, \text{cm}$.
An object is placed at $P$ in front of a spherical surface with refractive index $1.5$. Its image is formed at $Q$. If $PO = OQ$,find the distance $PO$. (in $R$)
Difficult
View SolutionParallel rays are incident on a transparent sphere along its one diameter. After refraction,these rays converge at the other end of this diameter. The refractive index of the sphere is:
Medium
View SolutionThe figure shows a transparent sphere of radius $R$ and refractive index $\mu$. An object $O$ is placed at a distance $x$ from the pole of the first surface so that a real image is formed at the pole of the exactly opposite surface. If an object is placed at a distance $R$ from the pole of the first surface,then the real image is formed at a distance $R$ from the pole of the second surface. The refractive index $\mu$ of the sphere is given by
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