Question

a) The focal length of a converging lens is 35 cm. An object is placed 100 cm in front of the lens. Describe the image.

b) The focal length of a converging lens is 35 cm. An object is placed 30 cm in front of the lens. Describe the image.

c) The focal length of a diverging lens is 35 cm. An object is placed 100 cm in front of the lens. Describe the image.

d) The focal length of a diverging lens is 35 cm. An object is placed 30 cm in front of the lens. Describe the image.

Answer 1

a)

here we have given that

lense type -converging lens - Convex lenses

focal length = 35 cm

object distance =  100 cm in front

the lens equation is -

1/v+1/u=1/f

on plugging the values we will get

1/v = -1/u +1/f = 1/35 +1/-100

therefore image distance will be v = 53.846 cm

nature is inverted, real image.

b.

here we have given that

lense type -converging lens - Convex lenses

focal length = 35 cm

object distance =   30 cm in front

the lens equation is -

1/v+1/u=1/f

on plugging the values we will get

1/v = -1/u +1/f = 1/35 +1/-30

therefore image distance will be v = -210 cm

nature is upright., a virtual image

c.

here we have given that

lense type -diverging lens - Concave lenses

focal length = 35 cm

object distance =  100 cm in front

the lens equation is -

1/v+1/u=1/f

on plugging the values we will get

1/v = -1/u +1-/f = 1/-35 +1/-100

therefore image distance will be v = -25.9259 cm

nature is virtual, upright image.

d.

here we have given that

lense type -diverging lens - Concave lenses

focal length = 35 cm

object distance = 30 cm in front

the lens equation is -

1/v+1/u=1/f

on plugging the values we will get

1/v = -1/u +1-/f = 1/-35 +1/-30

therefore image distance will be v = -16.15384 cm

nature is virtual, upright image.