Refraction of Light at Curved Surface

The value of $\cos \left(\theta \right)=\theta$ when paraxial approximation is done on the rays.

By paraxial approximation, the value of $\sin \left(\theta \right)=\_\_\_\_\_$.

Explain paraxial approximation.

Paraxial approximation is the big angle approximation used in different areas.

A real image of an object is formed at a distance of $20 \mathrm{c}\mathrm{m}$ from a lens. On putting another lens in contact with it, image is shifted $10 \mathrm{c}\mathrm{m}$ towards the combination. The power of the lens is _____ $\mathrm{D}$.

The radius of curvature of each surface of a convex lens of refractive index $1.5$ is $40 \mathrm{c}\mathrm{m}$. Calculate the power of this lens.

A real image of an object is formed at a distance of $20 \mathrm{c}\mathrm{m}$ from a lens. On putting another lens in contact with it, image is shifted $10 \mathrm{c}\mathrm{m}$ towards the combination. The power of the lens is _____ $\mathrm{D}$

An object is placed at distance of $0.06 \mathrm{m}$ from convex lens of focal length $0.10 \mathrm{m}$. Find the position of image.

A thin convex lens of glass $\left({\mathrm{n}}_{\mathrm{g}}=1.5\right)$ has power of $+5.0 \mathrm{D}$. When this lens is placed in a liquid of refractive index ${\mathrm{n}}_1$, it behaves like a concave lens of focal length $100 \mathrm{cm}.$Find value of ${\mathrm{n}}_1$.

A double convex lens of refractive index $1.47$ is immersed in a liquid. It behaves like a plane sheet. It means the refractive index of the liquid is-

A biconvex symmetrical lens made of glass of refractive index $1.56$ has radii of curvature of magnitude $20 \mathrm{c}\mathrm{m}$. Find the position of the image formed if an object is placed at a distance of $10 \mathrm{c}\mathrm{m}$ from this lens.

Two thin symmetrical lens of different nature have equal radii of curvature of all faces R = 20 cm. The lenses are put close together and immersed in water. The focal length of the system is 24 cm. The difference between refractive indices of the two lenses is …… $\times \frac{1}{9}$ Refractive index of water is $\frac{4}{3}$.

A person with normal near point $25 \mathrm{c}\mathrm{m}$ uses a compound microscope with an objective of focal length $8.0 \mathrm{m}\mathrm{m}$ and an eyepiece of focal length $2.5 \mathrm{c}\mathrm{m}$ to bring an object placed at $9.0 \mathrm{m}\mathrm{m}$ from the objective in sharp focus. The separation between two lenses and magnification respectively are

An object is placed at a distance of $1.5 \mathrm{m}$ from the screen and a convex lens is interposed between them. The magnification produced is $4$. What is the focal length of the lens.

A convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens. Then, it's focal length will:

A convex lens of focal length $0.2 \mathrm{c}\mathrm{m}$ and made of glass ${(}^{\mathrm{a}}{\mathrm{\mu }}_{\mathrm{w}}=1.5)$ is immersed in water ${(}^{\mathrm{a}}{\mathrm{\mu }}_{\mathrm{w}}=1.33)$. Find the change in the focal length of the lens.

A concave lens of glass, refractive index 1.5 has both surfaces of same radius of curvature R. On immersion in a medium of refractive index 1.75, it will behave as a