A 4.0-cm tall object is placed 70.0 cm from a diverging lens having a focal leng

Question

A 4.0-cm tall object is placed 70.0 cm from a diverging lens having a focal length of magnitude 35.0 cm. What are the nature and location of the image? The image is

virtual, 4.0 cm tall, and 20 cm from the lens on the side opposite the object. virtual, 2.0 cm tall, and 10 cm from the lens on the side opposite the object. virtual, 1.3 cm tall, and 23.3 cm from the lens on the same side of the object. real, 1.3 cm tall, and 23.3 cm from the lens on the same side of the object. real, 4.0 cm tall, and 20 cm from the lens on the side opposite the object.

Explanation / Answer

Option 3) virtual, 1.3 cm tall, and 23.3 cm from the lens on the same side of the object. is the correct answer

here is the explanation.

Given
object distance, u = 70 cm

focal length, f = -35 cm (for diverging lense f is negiative)

let v is object distance.

Apply

1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-35) - 1/70

v = -23.3 cm (image distance)

here the image is virtual.

maginification, m = -v/u

= -(-23.3)/70

= 0.3333

image height = m*object height

= 0.333*4

= 1.33 cm

so,

Option 3) virtual, 1.3 cm tall, and 23.3 cm from the lens on the same side of the object.

is the correct answer.

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