Chapter 1: Light - Reflection and Refraction

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Q
Light - Reflection and Reflection

Question:

An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.

Answer:

Here, the given lens is a convex lens.

Object Distance (u) = -25 cm

Focal length (f) = +10 cm

Height of object (h_o) = 5 cm

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

And, focal length (f) is taken positive because the lens is converging in nature.

Using the lens formula,

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

frac{1}{10} = frac{1}{v}-frac{1}{(-25)}

frac{1}{10} = frac{1}{v} + frac{1}{25}

frac{1}{10} -frac{1}{25} = frac{1}{v}

frac{1}{v} = frac{5-2}{50}

frac{1}{v} = frac{3}{50}

v=frac{50}{3}= 16.66 cm

The positive sign of v indicates that the image is formed on the other side of the convex lens at a distance of 16.66 cm. That is image is real in nature.

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{16.66}{-25} = -0.66

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

m = frac{h_i}{h_o}

-0.66 = frac{h_i}{5}

h_i = -0.66 	imes 5 = -3.3 cm

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

Therefore, the image is formed at a distance of 16.66 cm on the other side of the lens. The image formed is 3.3 cm tall and it is real and inverted.

The ray diagram given below shows the position, nature and size of the image.


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