Question

A converging lens of 6 cm focal length is mounted at a distance of 10 cm from a screen at right angles to the axis of the lens. A diverging lens of 12 cm focal length is then placed coaxially between the converging lens and the screen so that an image of an object 24 cm from the converging lens is formed in the screen. What is the distance between two lenses?​

Answer 1

A converging lens of 6 cm focal length is mounted at a distance of 10 cm from a screen at right angles to the axis of the lens. A diverging lens of 12 cm focal length is then placed coaxially between the converging lens and the screen so that an image of an object 24 cm from the converging lens is formed in the screen. What is the distance between two lenses?.

Answer 2

Answer:

The distance between the two lenses is 4 cm.

Explanation:

The focal length f of the converging lens = 6 cm

The object distance u from the converging lens = -24 cm

Let the image distance made by converging lens be v.

By lens equation,

1/f = 1/v - 1/u

1/6 = 1/v - 1/(-24)

v = 8 cm

In the figure, image forms at point A. This image is object for second diverging lens. Let the distance of diverging lens from this point be x and distance from the screen be t.

OA = 8 cm and AO' = t - x = 10 - 8 = 2 cm

The image distance for diverging lens is t and object distance is x. Focal length of diverging lens is = -12 cm.

By applying lens formula to diverging lens,

1/(-12) = 1/t - 1/x

But t = x+2

1/-12 = 1/(2+x) - 1/x

$24 = 2x + x^{2}$

$x^{2} + 2x - 24 = 0$

$(x-4)(x-6) = 0$

Now, x has two roots 4 and -6. But x can not be negative. So, x = 4 cm.

t = x + 2 = 6 cm

The distance of screen from converging lens is 10 cm and t = 6 cm.

So, the length between converging lens and diverging lens = 10 - 6 = 4 cm.