Question

A 2 cm long image is placed perpendicular to the principal axis of a convex lens of 10 cm focus distance. The distance of the object from the lens is 15 cm. is . Find the nature, position and size of the image.​

Answer 1

Answer:

given:

I=2cm

f=10cm

u=-15cm

lens formula

1/f=1/v + 1/u

1/10=1/v +1/-15

1/10+1/15=1/v

(3+2) /30=1/v

5/30=1/v

1/6=1/v

v=6cm(position of image)

nature of image- real and inverted

Answer 2

Answer:

the image is virtual,magnified, upright and on the opposite side of the lens as the object.

Explanation:

using the lens formula: 1/f=1/u+1/v

rearranging the above equation gives; v=uf/(f-u)

v= 15*10/(10-15) =-30cm the negative sign implies that:

1. the image is on the opposite side of lens as the object
2. the image is virtual (applying the real is positive sign convention). the image is also uprightsince all virtual images are upright.

the magnification of the object =v/u=height of image/height of object

           m=30cm/15cm=2

heght of image =magnification * height of object= 2cm*2=4cm

the image is larger than image thus, it is magnified.