The following figure shows a beam of light converging at point $P$. When a concave lens of focal length $16 \,cm$ is introduced in the path of the beam at a place shown by the dotted line such that $OP$ becomes the axis of the lens,the beam converges at a distance $x$ from the lens. The value of $x$ will be equal to (in $\,cm$)
- A$12$
- B$24$
- C$36$
- D$48$
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$A$ thin convex lens of refractive index $1.5$ has a focal length of $15\, cm$ in air. When the lens is placed in a liquid of refractive index $4/3$,its focal length will be......$cm$.
Medium
View SolutionParallel rays are incident on a thick plano-convex lens having radius of curvature $R$,refractive index $\mu$ and thickness $t$. When rays are incident on the plane surface,they converge at a distance $x$ from the plane surface. When rays are incident on the curved surface,they converge at a distance $y$ from the curved surface. Then:
Medium
View Solution$A$ convex lens of refractive index $1.5$ and focal length $f = 18 \ cm$ is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is $\alpha \times f$. The value of $\alpha$ is . . . . . . . (Refractive index of water $= 4/3$)
In the diagram shown,the lens is moving towards the object with a velocity $V \, m/s$ and the object is also moving towards the lens with the same speed. What is the speed of the image with respect to the earth when the object is at a distance $2f$ from the lens? ($f$ is the focal length.)
Difficult
View Solution$A$ convex lens of refractive index $1.5$ and focal length $18 \, cm$ in air is immersed in water. The change in focal length of the lens will be $........... \, cm$. (Given refractive index of water $= 4/3$)
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