Question

If an object is placed 50 cm in front of a concave mirror of 60 cm radius, what is the image size if object size is 15cm?​

Answer 1

Answer

Step-by-step explanation:

Given :-

R = -60 cm

u = -50 cm

Focal length = R/2

= -60/2

= -30 cm

By using mirrors formula,

The negative sign shows that image formed by mirror will be real and inverted .

Image distance from concave mirror is -75 cm.

Answer 2

Given :

Object distance : 50 cm.

Radius of curvature : 60 cm.

Height of object : 15 cm.

To find :

The size of the image.

Solution :

First we have to find the focal length of the mirror,

we know that,

» For a spherical mirror having small aperture, the principle focus lies exactly mid way between the pole and centre of curvature. So, the focal length of a spherical mirror is equal to the half of its radius of curvature.

if f is the focal length of a mirror and R is its radius of curvature, then f = R/2

by substituting the given values in the formula,

$\dashrightarrow \sf f = \dfrac{R}{2}$

$\dashrightarrow \sf f = \dfrac{60}{2}$

$\dashrightarrow \sf f = 30 \: cm$

Now using mirror formula that is,

» A formula which gives the relationship between image distance, object distance and focal length of a sperical mirror is known as the mirror formula .i.e.,

$\boxed{\bf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}}$

where,

• v denotes Image distance
• u denotes object distance
• f denotes focal length

By substituting all the given values in the formula,

$\dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$

$\dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{ - 50} = \dfrac{1}{ - 30}$

$\dashrightarrow\sf \dfrac{1}{v} - \dfrac{1}{50} = \dfrac{1}{ - 30}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1}{ - 30} + \dfrac{1}{50}$

$\dashrightarrow\sf \dfrac{1}{v} =\dfrac{ - 5 + 3}{150}$

$\dashrightarrow\sf \dfrac{1}{v} =\dfrac{ - 2}{150}$

$\dashrightarrow\sf \dfrac{1}{v} =\dfrac{ -1}{75}$

$\dashrightarrow\sf v = - 75 \: cm$

We know that,

» The linear magnification produced by a mirror is equal to the ratio of the image distance to the object distance with a minus sign and it is equal to the ratio of image height and object height. that is,

$\boxed{\bf m = \dfrac{h'}{h} = - \dfrac{v}{u}}$

where,

• h' denotes height of image
• h denotes object height
• v denotes image distance
• u denotes object distance

By substituting all the given values in the formula,

$\dashrightarrow\sf \dfrac{h'}{h} = - \dfrac{v}{u}$

$\dashrightarrow\sf \dfrac{h'}{15} = - \dfrac{ - 75}{ - 50}$

$\dashrightarrow\sf h' = - \dfrac{75 \times 15}{50}$

$\dashrightarrow\sf h' = - \dfrac{1125}{50}$

$\dashrightarrow\sf h' = - 22.5 \: cm$

Thus, the size of the image is 22.5 cm.