Question

Find the position and magnification of the image of an object placed at distance of 8.0 cm in front of a convex lens of focal length 10.0 cm. Is the image erect or inverted ?​

Answer 1

Explanation:

Image distance, (v) = -40 cm

Nature : Virtual and Erect

Magnification, (m) = +5

Step-by-step explanation:

Given,

Focal length of the lens, (f) = 10 cm

Object distance, (u) = - 8 cm

Let the distance of the image be 'v'. Then, we know that,

1/f = 1/v - 1/u

=> 1/v = 1/f + 1/u

=> 1/v = 1/10 + (-1) / 8

=> 1/v = 8/80 - 10/80

=> 1/v = -2/80

=> 1/v = -1/40

=> v = -40 cm

Since, the distance of image is negative hence it is on the side of object.

So the image is erect.

Nature : Virtual and Erect.

We know that,

Magnification, (m) = v/u

m = -40/(-8)

m = +5

Hence, the image is magnified to +5 times.