Chapter 1: Light - Reflection and Refraction

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Q
Light - Reflection and Reflection

Question:

An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.

Answer:

Here, the given mirror concave.

Object Distance (u) = -27 cm

Focal length (f) = -18 cm

Height of object (h_o) = 7 cm

Image Distance (v) = ?

Height of image (h_i) = ?

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

By using the mirror formula

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

frac{1}{-18} = frac{1}{v}+frac{1}{(-27)}

frac{1}{-18} = frac{1}{v} - frac{1}{27}

frac{1}{-18} +frac{1}{27} = frac{1}{v}

frac{1}{v} = frac{-3 + 2}{54}

frac{1}{v} = frac{-1}{54}

v = -54 cm

The screen should be placed at a distance of 54 cm in front of the mirror to obtain a sharp image.

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{-54}{-27}

m= -2

Here, the magnification is negative and more than 1. The negative sign indicates that the image is inverted and value more than 2 indicates that the size of the image is larger than the object.

Now,

m = frac{h_i}{h_o}

-2 = frac{h_i}{7}

h_i = -2 	imes 7 = -14 cm

The negative sign of image height h_i indicates that the image is inverted.

Therefore, the image is real, inverted and enlarged in size.


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