Question

When the distance of an object from a concave mirror is decreased from 15cm to 9cm , the image gets magnified 3 times than in first case. Calculate the focal lenght of the mirror. ​

Answer 1

Answer:

The focal length of the concave lens is -6cm

Explanation:

Let us consider an object AB be placed at 15cm in front of a concave mirror then the distance is reduced to 9cm

• The focal length be f

So ,

• Object distance at first case ,

$u_{1} = - 15cm$

and

• the object distance in the second case when distance is reduced ,

$u_2 = - 9cm$

Formula to be used

Magnification ,

$m = \frac{ h_{2} }{h _{1}} = - \frac{v}{u} = \frac{f}{f - u} = \frac{f - v}{f }$

So calculating the magnification form first case

$m_1 = \frac{f}{f - u_1} \\ \implies m_1 = \frac{f}{f - ( - 15)} \\ \implies m_1 = \frac{f}{ f + 15}$

And magnification for second case

$m_2 = \frac{f}{f -u_2 } \\ \implies m_2 = \frac{f}{f - ( - 9)} \\ \implies m_2 = \frac{f}{f + 9}$

Since , the image get magnified 3 times in the second case than the first case so

$m_2 = 3m_1 \\ \implies \frac{f}{f + 9} = 3 \times \frac{f}{f + 15} \\ \implies \frac{1}{f + 9} = \frac{3}{f + 15} \\ \implies f + 15 = 3(f + 9) \\ \implies f + 15 = 3f + 27 \\ \implies f - 3f = 27 - 15 \\ \implies - 2f = 12 \\ \implies f = - 6cm$

Thus , the focal length of the given concave mirror is -6cm ( the '-' sign signify here that the focal length of the concave mirror is taken in front it )