Question

An object is placed at a distance of 0.8m from the combination of convex lens of focal length 0.4m and concave lens of focal length 0.5m. Find the image distance and linear magnification.

Answer 1

Explanation:

Object will be beyond Centre of curvature.

So,

Image will be between focus and centre of curvature.

It will be real, inverted and diminished (small sized) image.

C = 2f

f is the focus.

Now , u = -0.5 m

f = -0.2 m

-» u sin g the lens formula

1/-0.5 = 1/v - 1/-0.2

1/-0.5 × 0.1x = 1/x × 0.1x - 1/-0.2 × 0.1x

= 0.1/-0.5x = 0.1 + 0.1/0.2x

= -0.1/0.5x - 0.1/0.2x = 0.1 + 0.1/0.2x - 0.1/0.2x

= -0.2x - 0.1/0.2x = 0.1

= -0.2x - 0.5x = 0.1

= - 0.7x = 0.1

= -0.7x/-0.7 = 0.1/-0.7

= x = 0.1/0.7

x = 1/7

And then = 1/7/-0.5

-1/3.5 is the final answer.

Or -2/7

The image formed is diminished, Virtual and erect .

Answer 2

Answer:

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Explanation:

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