A concave mirror of focal length 10 cm forms | vedclass

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Question

(B) For a concave mirror, the magnification $m$ can be $\pm 2$ because the image can be real (inverted) or virtual (erect).The focal length $f = -10$

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(B) For a concave mirror, the magnification $m$ can be $\pm 2$ because the image can be real (inverted) or virtual (erect).The focal length $f = -10$ cm.$1$) For real image, $m = -2$:Using $m = \frac{f}{f-u}$, we have $-2 = \frac{-10}{-10-u} \Rightarrow -2 = \frac{10}{10+u} \Rightarrow -20 - 2u = 10 \Rightarrow 2u = -30 \Rightarrow u_1 = -15$ cm.$2$) For virtual image, $m = +2$:Using $m = \frac{f}{f-u}$, we have $2 = \frac{-10}{-10-u} \Rightarrow 2 = \frac{10}{10+u} \Rightarrow 20 + 2u = 10 \Rightarrow 2u = -10 \Rightarrow u_2 = -5$ cm.The distance between the two positions is $|u_1 - u_2| = |-15 - (-5)| = |-10| = 10$ cm.

Column-$I$	Column-$II$
$1$. $m = -2$	a. Convex mirror
$2$. $m = -1/2$	b. Concave mirror
$3$. $m = +2$	c. Real image
$4$. $m = +1/2$	d. Virtual image

$A$ concave mirror of focal length $10$ cm forms an image which is double the size of the object when the object is placed at two different positions. The distance between the two positions of the object is . . . . . . cm.

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The focal length of a concave mirror is $f$ and the distance from the object to the principal focus is $x$. The magnification will be:

Match the corresponding entries of Column-$I$ with Column-$II$. [Where $m$ is the magnification produced by the mirror]
Column-$I$ Column-$II$
$1$. $m = -2$ a. Convex mirror
$2$. $m = -1/2$ b. Concave mirror
$3$. $m = +2$ c. Real image
$4$. $m = +1/2$ d. Virtual image

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