(D) For a concave mirror,the magnification $m$ is given by $m = \frac{f}{f-u}$.For the first position,$u_1 = -25 \ cm$,so $m_1 = \frac{f}{f - (-25)} =

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(D) For a concave mirror,the magnification $m$ is given by $m = \frac{f}{f-u}$.For the first position,$u_1 = -25 \ cm$,so $m_1 = \frac{f}{f - (-25)} = \frac{f}{f+25}$.For the second position,the object is moved $15 \ cm$ farther,so $u_2 = -(25 + 15) = -40 \ cm$. Thus,$m_2 = \frac{f}{f - (-40)} = \frac{f}{f+40}$.Given $\frac{m_1}{m_2} = 4$,we substitute the expressions: $\frac{f/(f+25)}{f/(f+40)} = 4$.This simplifies to $\frac{f+40}{f+25} = 4$.$f + 40 = 4(f + 25) \implies f + 40 = 4f + 100$.$3f = -60 \implies f = -20 \ cm$.The focal length of a concave mirror is $20 \ cm$ (magnitude).

Class 12 Physics Ray Optics and Optical Instruments Spherical Mirror

Difficult

When an object is placed at a distance of $25 \ cm$ from a concave mirror,the magnification is $m_1$. The object is moved $15 \ cm$ farther away with respect to the earlier position and the magnification becomes $m_2$. If $m_1/m_2 = 4$,the focal length of the mirror is $....... \ cm$ (Assume image is real and $m_1, m_2$ are numerical values).

A
$10$
B
$30$
C
$15$
D
$20$

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Ray Optics and Optical Instruments

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Spherical Mirror

The cross-section of a reflecting surface is represented by the equation $x^{2}+y^{2}=R^{2}$ as shown in the figure. $A$ ray travelling in the positive $x$ direction is directed toward the positive $y$ direction after reflection from the surface at point $M$. The coordinate of the point $M$ on the reflecting surface is

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$A$ point object is between the Pole and Focus of a concave mirror,and moving away from the mirror with a constant speed. Then,the velocity of the image is:

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$A$ straight metal rod of length $6 \ cm$ is placed along the principal axis of a concave mirror of focal length $9 \ cm$ such that the end of the rod closer to the mirror is at a distance of $15 \ cm$ from the pole of the mirror. The length of the image of the rod is (in $cm$)

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An object is placed in front of a concave mirror of focal length $f$. $A$ virtual image is formed with a magnification of $2$. To obtain a real image of the same magnification,the object has to be moved by a distance of:

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$A$ mobile phone lies along the principal axis of a concave mirror,as shown in the figure. Show by a suitable diagram,the formation of its image. Explain why the magnification is not uniform. Will the distortion of the image depend on the location of the phone with respect to the mirror?

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$A$ point object is between the Pole and Focus of a concave mirror,and moving away from the mirror with a constant speed. Then,the velocity of the image is:

$A$ straight metal rod of length $6 \ cm$ is placed along the principal axis of a concave mirror of focal length $9 \ cm$ such that the end of the rod closer to the mirror is at a distance of $15 \ cm$ from the pole of the mirror. The length of the image of the rod is (in $cm$)

An object is placed in front of a concave mirror of focal length $f$. $A$ virtual image is formed with a magnification of $2$. To obtain a real image of the same magnification,the object has to be moved by a distance of:

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