(B) Given: Object distance $u = -9 \, cm$,Focal length $f = 10 \, cm$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Substituting t

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(B) Given: Object distance $u = -9 \, cm$,Focal length $f = 10 \, cm$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Substituting the values: $\frac{1}{10} = \frac{1}{v} - \frac{1}{-9} \Rightarrow \frac{1}{v} = \frac{1}{10} - \frac{1}{9} = \frac{9-10}{90} = -\frac{1}{90}$.Thus,the image distance $v = -90 \, cm$.The magnification $m$ is given by $m = \frac{v}{u}$.$m = \frac{-90}{-9} = 10$.Therefore,the magnification produced by the lens is $10$.

Class 12 Physics Ray Optics and Optical Instruments Refraction by Lenses

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$A$ card sheet divided into squares each of size $1 \, mm^2$ is being viewed at a distance of $9 \, cm$ through a magnifying glass (a converging lens of focal length $10 \, cm$) held close to the eye. What is the magnification produced by the lens?

A
$5$
B
$10$
C
$15$
D
$20$

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Class 12 Questions

Class 12

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Ray Optics and Optical Instruments

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Refraction by Lenses

In the figure,an air lens of radii of curvature $10 \ cm$ ($R_1 = 10 \ cm$,$R_2 = -10 \ cm$) is cut in a cylinder of glass $(\mu = 1.5)$. The focal length and the nature of the lens is:

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For a concave lens of focal length $f$,the relation between object distance $u$ and image distance $v$ from its pole can best be represented by ($u = v$ is the reference line):

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The refractive index of a glass convex lens is $1.5$. The radius of curvature of each of the two surfaces of the lens is $40 \ cm$. The ratio of the power of the lens when immersed in a liquid of refractive index $1.25$ to that when placed in air is:

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The distance between the object and the screen is $100 \, cm$. $A$ lens produces an image on the screen when it is placed at either of two positions $40 \, cm$ apart. The power of the lens is (approximately): (in $, D$)

Medium

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For a convex lens,the distance of the object is taken on the $X$-axis and the distance of the image is taken on the $Y$-axis. The nature of the graph so obtained is:

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$A$ card sheet divided into squares each of size $1 \, mm^2$ is being viewed at a distance of $9 \, cm$ through a magnifying glass (a converging lens of focal length $10 \, cm$) held close to the eye. What is the magnification produced by the lens?

Similar Questions

In the figure,an air lens of radii of curvature $10 \ cm$ ($R_1 = 10 \ cm$,$R_2 = -10 \ cm$) is cut in a cylinder of glass $(\mu = 1.5)$. The focal length and the nature of the lens is:

For a concave lens of focal length $f$,the relation between object distance $u$ and image distance $v$ from its pole can best be represented by ($u = v$ is the reference line):

The refractive index of a glass convex lens is $1.5$. The radius of curvature of each of the two surfaces of the lens is $40 \ cm$. The ratio of the power of the lens when immersed in a liquid of refractive index $1.25$ to that when placed in air is:

The distance between the object and the screen is $100 \, cm$. $A$ lens produces an image on the screen when it is placed at either of two positions $40 \, cm$ apart. The power of the lens is (approximately): (in $, D$)

For a convex lens,the distance of the object is taken on the $X$-axis and the distance of the image is taken on the $Y$-axis. The nature of the graph so obtained is:

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