When two lenses are used in combination, the first one forms an image that then

Question

When two lenses are used in combination, the first one forms an image that then serves as the object for the second lens. The magnification of the combination is the ratio of the height of the final image to the height of the object. A 1.20 cm -tall object is 60.0 cm to the left of a converging lens of focal length 40.0 cm . A second converging lens, this one having a focal length of 60.0 cm , is located 300 cm to the right of the first lens along the same optic axis.

I1 is now the object for the second lens. Find the location and height of the image produced by the second lens. This is the final image produced by the combination of lenses.

Explanation / Answer

Take the case of the first lens –

Use the universal lens formula –

1/u+1/v=1/f.

u = 60, f = 40
1/v = 1/40 - 1/60

=> 1/v = 0.0083

=> v = + 120 cm
It means 120 cm to the right of the 1st lens.
And magnification = v/u = 120 / 60 = 2
Therefore, image height = 2 x 1.20 = +2.40 cm (erect).

Now, take the case of the second lens -

1/u+1/v=1/f.
u = 300 – 120 = 180, f = -60.
1/v=1/-60-1/180

=> 1/v = - 0.0222

=> v = - 45.0 cm
Means, image is to the left of the 2nd lens, and the image is virtual.
Magnification=v/u = -45.0 / 180 = -0.25
So, the final image height hv = -0.25 x 2.40 = -0.60 cm (inverted).

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