Question

the object is placed in front of concave with radius of curvature of 20cm determine the image distance if the object is 15



please explain correctly and perfectly with step step please this is my humble request ​

Answer 1

Step-by-step explanation:

Question :

The object is placed in front of concave with radius of curvature of 20cm determine the image distance if the object is 15.

Answer :

Distance of image is 30 cm

Explanation :

Given :

• u = -15 cm
• R = 20 cm

To Find :

• Image distance

Formula Used :

• f = R/2
• 1/f = 1/v+1/u

Step by step explanation :

$\boxed{ \bf{ \purple{f = R/2}}}$

$\implies \tt \: f = \frac{20}{2} = 10 \: cm \:$

By using mirror formula :

$\mapsto \bf \: 1/v=1/f-1/u$

$\mapsto \bf \: \frac{1}{v} = \frac{1}{ - 10} - \frac{1}{ - 15}$

$\mapsto \bf \: \frac{1}{v} = \frac{1}{15} - \frac{1}{10}$

$\mapsto \bf \: \frac{1}{v} = \frac{2 - 3}{30}$

$\mapsto \bf \: \frac{1}{v} = \frac{ - 1}{30}$

$\mapsto \bf \: v \: = - 30 \: cm$

The negative sign indicates that the image is formed at same side as that of object .

Image distance is 30 cm .

Nature of image is real , inverted and magnified .

Answer 2

Given :

In concave mirror,

Radius of curvature : 20 cm.

Object distance : 15 cm.

To find :

The image distance of the mirror.

Solution :

first we have to find the focal length,

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,

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

$\dashrightarrow \sf f = \dfrac{20}{2}$

$\dashrightarrow \sf f = 10 \: cm$

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,

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

$\dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{ - 15} = \dfrac{1}{ - 10}$

$\dashrightarrow\sf \dfrac{1}{v} - \dfrac{1}{15} = - \dfrac{1}{10}$

$\dashrightarrow\sf \dfrac{1}{v} = - \dfrac{1}{10} + \dfrac{1}{15}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{ - 3 + 2}{30}$

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

$\dashrightarrow\sf v = - 30 \: cm$

Thus, the image distance is -30 cm.