Question

5 centimetre tall object is placed perpendicular to the principal axis of convex lens of focal length is 10 centimetre the distance of the object from the lens is 15 cm find the nature position and size of the image.also find it magnification

Answer 1

Answer:

Position of the image = 30 cm

Magnification of the image = - 2

The image is real, inverted and magnified.

Explanation:

ho = 5 cm

f = 10 cm

u = - 15 cm

Lens formulae : 1 /v - 1 /u = 1 /f

1 /v = 1 /f + 1/u

      = 1 /10 - 1 /15

      = 5 /150 = 1 /30

v = 30 cm (Position of the image)

Magnification : m = v /u

m = 30 / -15

   = - 2 (The image is real, inverted and magnified)

     

Answer 2

Answer: The position of image = 30 cm and Magnification of image = -2. The image is real, inverted and magnified.

Explanation:

Given:

Initial height Ho = 5 cm

Focal length = 10 cm

object distance u = 15 cm

Find the position of image v:

Lens formula = $\frac{1}{v} = \frac{1}{f} + \frac{1}{u}\\\frac{1}{v} = \frac{1}{10} - \frac{1}{15}$

$v = \frac{150}{15-10} = \frac{150}{5} = 30cm$

Find the magnification

m = $\frac{v}{u} = \frac{30}{-15}=-2$

Therefore, The position of image = 30 cm and Magnification of image = -2. The image is real, inverted and magnified.

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