Chapter 1: Light - Reflection and Refraction

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Q
Light - Reflection and Reflection

Question:

An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.

Answer:

Focal length (f) = 15 cm

Object distance (u) = -10 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.

And, focal length (f) is taken positive according to the sign convention, as the mirror is convex.

By using the mirror formula, we get

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

frac{1}{15} = frac{1}{v}+frac{1}{(-10)}

frac{1}{15} +frac{1}{10} = frac{1}{v}

frac{1}{v} = frac{2+3}{30}

frac{1}{v} = frac{5}{30}

v = 6 cm

The positive sign of v indicates that the image is formed behind the convex mirror at a distance of 6 cm. That is the image formed is virtual in nature.

Now to find the nature of the image we will find the magnification.

m= -frac{v}{u}

m= -frac{6}{-10} = 0.6

The positive value of m indicates that the image formed is erect and the value of m is less than 1 indicates the image formed is diminished in size.

Therefore, the image formed at a distance of 6 cm behind the mirror. The image formed is virtual, erect and diminished in size.


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