Answer  Hint:In order to solve this question, you must be aware of the relationship between the distance of an image (v), the distance of an object (u), and the focal length (f) of the mirror. This is given by a formula known as mirror formula. It is applicable for convex as well as mirrors. Complete step by step answer: A mirror is a reflective surface that bounces off light, producing either a real image or a virtual image. When an object is placed in front of a mirror, the image of the same object is seen in the mirror. The object is the source of the incident rays and the image is formed by the reflected rays. Based on the interaction of light, the images are classified as either a real image or a virtual image. A real image occurs when the light rays actually intersect while virtual images occur due to the apparent divergence of light rays from a point.(a) Given: magnification, $m = - 3$Object distance, $u = - 20$ cmUsing magnification formula,$m = \dfrac{{ - v}}{u}$$\Rightarrow - 3 = \dfrac{{ - v}}{{ - 20}}$$\Rightarrow v = - 60$ cmUsing mirror formula,$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$$\Rightarrow \dfrac{1}{f} = \dfrac{1}{{ - 60}} + \dfrac{1}{{ - 20}}$$\therefore f = - 15$ cm(b) To find the distance of the object from the mirror, when magnification given is 3Use magnification formula$m = \dfrac{{{h_i}}}{{{h_o}}}$$\Rightarrow m = \dfrac{{ - v}}{u}$$\Rightarrow 3 = \dfrac{{ - v}}{u}$$\Rightarrow v = - 3u$Using mirror formula,$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$$\Rightarrow \dfrac{1}{{ - 3u}} + \dfrac{1}{u} = \dfrac{1}{{ - 15}}$$\therefore u = - 10$ cmNote:When a spherical mirror surface is painted outwards and its inner surface is reflecting then it is known as a Concave mirror. Concave mirrors are used as a reflector in the headlights of cars and in search of light. (a) Given: ”‹Distance of the object from the mirror, $u$ = $-$20 cm Magnification, $m$ = $-$3 cm To find: Focal length of the mirror $(f)$. Solution: From the magnification formula, we know that- $m=-\frac{v}{u}$ Substituting the given values in the magnification formula we get- $-3=-\frac{v}{(-20)}$ $-3=\frac{v}{20}$ $v=-60cm$ Thus, the distance of the image, $v$ is 60 cm, and the negative sign implies that the image forms in front of the mirror (on the left). Now, from the mirror formula, we know that- $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$ Substituting the given values in the mirror formula we get- $\frac{1}{f}=\frac{1}{(-60)}+\frac{1}{(-20)}$ $\frac{1}{f}=-\frac{1}{60}-\frac{1}{20}$ $\frac{1}{f}=\frac{-1-3}{60}$ $\frac{1}{f}=-\frac{4}{60}$ $\frac{1}{f}=-\frac{1}{15}$ $f=-60cm$ Thus, the focal length, $f$ is 60 cm, and the negative sign implies that it is in front of the mirror (on the left). (b) If the image is virtual and 3 times magnified Given: Focal length of the mirror, $f$ = $-$15 cm Magnification, $m$ = 3 cm ”‹Distance of the object from the mirror, $u$ = $-$20 cm To find: Position or the distance of the object from the mirror, $(u)$. Solution: From the magnification formula, we know that- $m=-\frac{v}{u}$ Substituting the given values in the magnification formula we get- $3=-\frac{v}{u}$ $v=-3u$ Thus, the distance of the image, $v$ is 3u, and the negative sign implies that the image forms in front of the mirror (on the left). Now, from the mirror formula, we know that- $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$ Substituting the given values in the mirror formula we get- $\frac{1}{(-15)}=\frac{1}{(-3u)}+\frac{1}{u}$ $\frac{1}{-15}=-\frac{1}{3u}+\frac{1}{u}$ $\frac{1}{-15}=\frac{-1+3}{3u}$ $\frac{1}{-15}=\frac{2}{3u}$ $3u=2\times {(-15)}$ $u=\frac{-30}{3}$ $u=-10cm$ Thus, the object should be placed at a distance of 10 cm to get a virtual image three times the height of the object. The negative sign with the object implies that the object is placed in front of the mirror (on the left). |
When an object is placed at 20 cm from a concave mirror a real image magnified three times is formed?
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