Question

a concave lens has a focal length of 10cm an object 2.5 cm high is placed 30 cm from the lens. Determine the position and size of the image​

Answer 1

Answer:

कनकसययमकमस मगन न नगननग ग ुगम. मगमकम मकमकम मग. ीकियितितमकककवतप पद

Explanation:

लतपदतपयपपवदव ककवनन

Answer 2

Given:-

• Focal length ,f = -10cm

• Object height ,ho = 2.5 cm

• Object distance ,u = -30 cm

To Find:-

• The position and size of the image

Solution:-

By using lens Formula

• 1/v - 1/u = 1/f

Substitute the value we get

→ 1/v = 1/f +1/u

→ 1/v = -1/10 + 1/(-30)

→ 1/v = -1/10 -1/30

→ 1/v = -3-1/30

→ 1/v = -4/30

→ v = -30/4

→ v = -7.5 cm

Therefore, the image is formed at a distance of 7.5 cm on the same side of the object. The image formed is Virtual and Point Sized.

Now , calculating the Magnification

• magnification ,m = v/u = hi/ho

→ m = -7.5/-30 = hi/2.5

→ 7.5/30 = hi/2.5

→ hi = 2.5×7.5/30

→ hi = 18.75/30

→ hi = 0.62 cm

Therefore ,the height of image is smaller than the object . Height of image is 0.62 cm.