(B) Given: Focal length $f = -10 \, cm$,Object distance $u = -25 \, cm$,Area of object $A_{obj} = (3.0 \, cm)^2 = 9 \, cm^2$.Using the mirror formula,

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(B) Given: Focal length $f = -10 \, cm$,Object distance $u = -25 \, cm$,Area of object $A_{obj} = (3.0 \, cm)^2 = 9 \, cm^2$.Using the mirror formula,$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,we find the magnification $m = \frac{f}{f - u}$.$m = \frac{-10}{-10 - (-25)} = \frac{-10}{15} = -\frac{2}{3}$.The area of the image is given by $A_{image} = m^2 \times A_{obj}$.$A_{image} = \left( -\frac{2}{3} \right)^2 \times 9 = \frac{4}{9} \times 9 = 4 \, cm^2$.

Class 12 Physics Ray Optics and Optical Instruments Spherical Mirror

Difficult

$A$ square wire frame of side length $3.0 \, cm$ is placed $25 \, cm$ away from a concave mirror of focal length $10 \, cm$. The center of the wire frame is on the axis of the mirror,with two of its sides perpendicular to the axis. The area of the image will be $..... \, cm^2$.

A
$3.5$
B
$4$
C
$2.9$
D
$4.6$

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Ray Optics and Optical Instruments

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Spherical Mirror

$A$ boy stands straight in front of a mirror at a distance of $30 \ cm$ away from it. He sees his erect image whose height is $\frac{1}{5}$ of his real height. The mirror he is using is

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An object of height $5 \, cm$ is placed at a distance of $1 \, m$ from a concave mirror of radius of curvature $20 \, cm$. What will be the height of the image in $cm$?

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If the distance between an object and its two times magnified virtual image produced by a curved mirror is $15 \,cm$, the focal length of the mirror must be:

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$A$ rod of length $10 \text{ cm}$ lies along the principal axis of a concave mirror of focal length $10 \text{ cm}$ as shown in the figure. The length of the image is . . . . . . $\text{cm}$.

DifficultJEE MAIN 2026

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Define the following for curved mirrors: $(1)$ Radius of curvature,$(2)$ Centre of curvature,$(3)$ Pole,$(4)$ Principal axis,$(5)$ Aperture,$(6)$ Principal focus,$(7)$ Focal plane,$(8)$ Focal length,$(9)$ Paraxial rays.

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$A$ square wire frame of side length $3.0 \, cm$ is placed $25 \, cm$ away from a concave mirror of focal length $10 \, cm$. The center of the wire frame is on the axis of the mirror,with two of its sides perpendicular to the axis. The area of the image will be $..... \, cm^2$.

Similar Questions

$A$ boy stands straight in front of a mirror at a distance of $30 \ cm$ away from it. He sees his erect image whose height is $\frac{1}{5}$ of his real height. The mirror he is using is

An object of height $5 \, cm$ is placed at a distance of $1 \, m$ from a concave mirror of radius of curvature $20 \, cm$. What will be the height of the image in $cm$?

If the distance between an object and its two times magnified virtual image produced by a curved mirror is $15 \,cm$, the focal length of the mirror must be:

$A$ rod of length $10 \text{ cm}$ lies along the principal axis of a concave mirror of focal length $10 \text{ cm}$ as shown in the figure. The length of the image is . . . . . . $\text{cm}$.

Define the following for curved mirrors: $(1)$ Radius of curvature,$(2)$ Centre of curvature,$(3)$ Pole,$(4)$ Principal axis,$(5)$ Aperture,$(6)$ Principal focus,$(7)$ Focal plane,$(8)$ Focal length,$(9)$ Paraxial rays.

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