Question

A real image obtained by a concave mirror is 4 times begger than the object. If the object is displaced by 3 cm away from the mirror, the image size becomes 3 times the object size. Find the focal length of the mirror.

Answer 1

Answer:

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Answer 2

Answer:

-4.21 cm.

Explanation:

The magnification of the mirror is defined in two ways:

(1) In terms of the ratio of image height $h_i$ to the object height $h_o$ as

$m=\dfrac{h_i}{h_o}.$

(2) In terms of ratio of image distance $v$ to the object distance $u$ from the mirror as

$m=-\dfrac vu$

When the image obtained by a concave mirror is 4 times bigger than the object.

$h_i=4h_o$

The magnification of the mirror is given by

$m=\dfrac {h_i}{ h_o} = \dfrac{4h_o}{h_o} = 4\\4=-\dfrac vu \\\Rightarrow v=-4u.$

Using mirror equation,

$\dfrac 1f=\dfrac 1v+\dfrac 1u=\dfrac{1}{-4u}+\dfrac 1u=\dfrac{-1+4}{4u}=\dfrac {3}{4u}\\\Rightarrow f=\dfrac 43 u.\\u=\dfrac{3f}4\ \ \ \ \ .............\ \ (1).$

When the object is displaced by 3 cm away from the mirror, the image size becomes 3 times the object size.

Let the new object and image distances be $u'$ and $v'$ respectively.

Now,

$h_i = 3h_o\\m=\dfrac{h_i}{h_o} = \dfrac{3h_o}{h_o} = 3\\\therefore 3=-\dfrac {v'}{u'}\\v'=-3u'$

Given,

$u'=u+3$

Using mirror equation,

$\dfrac 1f=\dfrac 1{v'}+\dfrac 1{u'}=-\dfrac 1{3u'}+\dfrac 1{u'}=\dfrac {-1+3}{3u'}=\dfrac{2}{3u'}=\dfrac{2}{3(u+3)}$

Using (1),

$f=\dfrac{2}{3(3+u)}=\dfrac{2}{3+\dfrac {3f}4}\\f\left(9+\dfrac {9f}4 \right)=2\ \ \ \text{(Using (1))} \\9f+\dfrac {9f^2}{4} =2\\9f^2+36f-8=0\\\Rightarrow f=\dfrac{-36\pm\sqrt{36^2-4\cdot 9\cdot (-8)}}{2\times 9}=\dfrac{-36\pm 39.80}{18}\\=-4.21\ cm. \ \ \ \text{or} \ \ \ \ +0.21\ cm.$

The focal length of the concave mirror is always negative therefore, the focal length of the mirror is -4.21 cm.