Question

A converging lens with a focal length of 12.0cmforms a virtual image 8.00mm tall, 17.0cm to the right of the lens.

Part A

Determine the position of the object.

s =

  cm  

Part B

Determine the size of the object.

|y| =

  cm  

Part C

Is the image erect or inverted?

Is the image erect or inverted?

The image is erect.

The image is inverted.

Part D

Are the object and image on the same side or opposite sides of the lens?

Are the object and image on the same side or opposite sides of the lens?

on the same side

on opposite sides

Answer 1

Part A.

Using lens equation for converging lens:

1/f = 1/u + 1/v

u = image distance = -17.0 cm (-ve sign since image is on the right side of lens)

f = focal length of lens = 12.0 cm, So

1/u = 1/f - 1/v

1/u = 1/12.0 - 1/(-17.0)

u = 17.0*12.0/(17.0 + 12.0) = 7.03 cm

u = object distance = 7.03 cm

Part B.

Magnification is given by:

M = -v/u = hi/ho

ho = height of object = 8.00 mm

So, hi = height of image = ?

ho = hi*(-u/v)

ho = 8.00*(-7.03/(-17.0))

ho = 3.31 mm

ho = 0.33 cm = Height of object

Part C.

m = -v/u = -(-17.0)/7.03 = +2.42

Since m > 0, So image is erect

Part D.

Since image is virtual, So object and image are on the same side.

Let me know if you've any query.