Question

The focal length of a concave mirror is 'f ' and the distance of the object from the principal focus is 'a'. The magnitude of magnification obtained will be :

Answer 1

Answer:

Give that

focal length =f

now

-1/f = 1/v+1/4

Aceording to question

object distant, 4=-(f+p)

so,

-1/f=1/v+(-1/f+p)

or, v=-f (f+p)/p

Now the ratio of the size of the image to the size of the object is given by

T/o=V/u

i/o=f(f+p)/p(f+p)

i/o = f/p

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Answer 2

Answer:

The magnitude of magnification obtained will be $\mathrm{\frac{f}{a}}$.

Explanation:

We will use the mirror formula to solve this question.

$\mathrm{\frac{1}{f}=\frac{1}{v}+\frac{1}{u}}$           (1)

Where,

f=focal length of the mirror

v=image distance from the mirror

u=object distance from the mirror

From the question we have,

The focal length of the concave mirror= -f

The object distance from the mirror=-(f+a) (as the distance of the object from the principal focus is 'a')

By inserting the required values in equation (1) we get;

$\mathrm{\frac{1}{-f}=\frac{1}{v}+\frac{1}{-(a+f)}}$

$\mathrm{\frac{-1}{f}=\frac{1}{v}-\frac{1}{(a+f)}}$

$\mathrm{\frac{1}{v}=\frac{-1}{f}+\frac{1}{(a+f)}}$

$\mathrm{v=\frac{-(f+a)\times f}{a} }$       (2)  (-ve sign indicates the image is real)

And the magnification is calculated as,

$\mathrm{m=\frac{-v}{u} }$          (3)

Inserting the value of "v" and "u" in equation (3) we get;

$\mathrm{m=\frac{-(\frac{-(f+a)\times f}{a})}{-(f+a)} }$

$\mathrm{m=\frac{-f}{a} }$      (3)       (the magnitude will be positive)

So, the magnitude of magnification obtained will be $\mathrm{\frac{f}{a}}$.

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