Question

A concave mirror with radius 6m object distance 15m and height of the image 3m. Find the focal length, image distance and magnification​

Answer 1

Answer:

f= -3m

v= - 2.5m

m= -1.66

Explanation:

refer the attached image for solution

Answer 2

Given :

In concave mirror,

Radius of curvature : 6 m

Object distance : 15 m

Height of the image : 3 m

To find :

The focal length, image distance and the magnification of the image.

Solution :

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,

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

$\dashrightarrow \sf f = \dfrac{6}{2}$

$\dashrightarrow \sf f = 3 \: m$

Thus, the focal length is 3 m.

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,

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

$\dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{ - 15} = \dfrac{1}{ - 3}$

$\dashrightarrow\sf \dfrac{1}{v} - \dfrac{1}{15} = \dfrac{1}{ - 3}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1}{ - 3} + \dfrac{1}{15}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{ -5 + 1}{15}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{ -4}{15}$

$\dashrightarrow\sf \dfrac{1}{v} = - \dfrac{15}{4}$

$\dashrightarrow\sf v = - 3.75 \: m$

Thus, the position of the image is -3.75 m.

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

$\dashrightarrow\bf m = - \dfrac{v}{u}$

By substituting all the given values in the formula,

$\dashrightarrow\sf m = - \dfrac{ - 3.75}{ - 15}$

$\dashrightarrow\sf m = - 0.25 \:$

Thus, the magnification of the image is -0.25