Chapter 1: Light - Reflection and Refraction

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Light - Reflection and Reflection


An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.


Here, the given mirror convex.

Object Distance (u) = -20 cm

Radius of curvature (R) = +30 cm

Height of object ({h}{o}) = 5 cm

Image Distance (v) = ?

Object distance (u) is taken negative as their distances from the pole of the mirror is measured opposite to the direction of the incident light.

Using the relation between R and f:

R = 2f

We get,

f = frac{R}{2}

f = frac{30}{2}

f = 15 cm

Now, using the mirror formula

frac{1}{f} = frac{1}{v}+frac{1}{u}

frac{1}{15} = frac{1}{v}+frac{1}{(-20)}

frac{1}{15} = frac{1}{v} - frac{1}{20}
frac{1}{15} +frac{1}{20} = frac{1}{v}

frac{1}{v} = frac{4-3}{60}

frac{1}{v} = frac{7}{60}

v=frac{60}{7}= 8.57 cm

The positive sign of v indicates that the image is formed on the other side of the mirror at a distance of 8.57 cm. That is the image formed is virtual.

Now, magnification

m = frac{Heightquad of quad image (h_i)}{ Heightquad of quad object (h_o)} = frac{sizequad of quad image (v)}{sizequad of quad object (u)}

m=frac{h_i}{h_o}= -frac{v}{u}

m= -frac{v}{u}

m= -frac{8.57}{-20} = 0.428

Here, the magnification is positive and less than 1. The positive sign indicates that the image is erect and value less than 1 indicates that the size of the image is smaller than the object.m = frac{h_i}{h_o}
0.428 = frac{h_i}{5}

h_i = 0.428 	imes 5 = 2.14 cm

The positive sign of image height h_i indicates that the image is erect.

Therefore, the image is formed at a distance of 0.428 cm on the other side of the mirror. The image formed is 2.14 cm tall and it is virtual, erect and smaller in size.

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