Question

the focal length of a concave mirror is "f" and the distance of the object from the principal focus is "p". then the ratio of the size of the image to the size of the object is given by ???​

Answer 1

$\bold{\underline{\underline{Answer:}}}$

Ratio of the size of the image to the size of the object = $\bold{\dfrac{f}{p}}$

$\bold{\underline{\underline{Step\:by\:step\:explanation:}}}$

Given :

• The focal length of a concave mirror is " f "
• The distance of the object from the principal focus is "p"

To find :

• The ratio of the size of the image to the size of the object.

Solution :

We will use the mirror formula to solve out this question.

Since the mirror is a concave mirror it's focal length will be negative.

•°• Focal length = - f

Object distance (u) = p + f

Since the object is always placed on the LHS it is therefore negative.

•°• u = -f + p

Object distance = - f + p ---> (1)

Formula :

$\bold{\large{\boxed{\rm{\red{\dfrac{1}{-f}\:=\:{\dfrac{1}{v}\:+\:{\dfrac{1}{u}}}}}}}}$

Block in the values,

$\implies$$\bold{\dfrac{1}{-f}\:=\:{\dfrac{1}{v}\:+\:{\dfrac{1}{-f+p}}}}$

$\implies$$\bold{v\:=\:{\dfrac{-f(f+p)}{p}}}$

Let image be $\bold{I}$

Let object be $\bold{O}$

Ratio = $\bold{I:O}$

Magnification is given by the formula,

$\implies$$\bold{\dfrac{I}{O}}$ = $\bold{\dfrac{v}{u}}$

Block in the values,

$\implies$$\bold{\dfrac{I}{O}}$ = $\bold{\dfrac{-f(f+p)}{p(f+p)}}$

$\implies$$\bold{\dfrac{I}{O}}$ = $\bold{\dfrac{-f}{p}}$

Considering only the magnitude ,

$\implies$$\bold{\dfrac{I}{O}}$ = $\bold{\dfrac{f}{p}}$

•°• The ratio of the size of the image to the size of the object is given by, $\bold{\dfrac{I}{O}}$ = $\bold{\dfrac{f}{p}}$

Answer 2

Answer:

Explanation:

focal length=f

Distance of object from lens=p=x+f

then q=?

1/f=1/p+1/q

1/f-1/p=1/q

1/f-1/x+f=1/q

x/f(x+f)=1/q

=>q=f(x+f)/x

Now MagnificatiOn i.e

Size of image/size of object=q/p

={f(x+f)/x}÷(x+f)

=f(x+f)/x(x+f)

=f/x.