I\'m struggling on these questions, please show any and all work so I can better

Question

I'm struggling on these questions, please show any and all work so I can better understand! Thank you in advance!

Part A:

A converging lens is equiconvex, i.e., the radius of curvature for the two sides is the same. If the radius of curvature of each side is 0.25m, and the lens is made of glass of refractive index 1.5...

1) what is the focal length of the lens?

Part B:

The lens is used in a slide projector, where the lens and the screen can be moved, but the position of the object is fixed (with respect to the ground). Initially, the distance from the lens to the object is 0.5m.

2) Where should the screen be placed to be able to see a sharp image?

3) If the object is 0.05m tall, how big will the image be? Will it be upright or inverted?

4) The screen is now moved farther away from the lens. How should the lens be moved for the image to be sharp again? (Towards the object, away from the object, or not moved at all?) Explain.

Explanation / Answer

(A) Using lens maker's formula,

1/f = (n - 1)[1 /R1 - 1/R2]

1/f = (1.5 - 1)[ 1/0.25 - 1/(-0.25)]

f = 0.25 m ........Ans

(b) (2) Object distance, p = 0.50 m

Applying lens equation,

1/f = 1/p + 1/q

1/0.25 = 1/0.5 + 1/q

q = 0.50 m

so screen have to placed at distance of 0.50 m behind the lens.

(3) m = - q / p = hi / ho

- 0.50 / 0.50 = hi / 0.05

hi = - 0.05 m

image will be of same size.(0.04 m tall)

and inverted

(4) if screen is moved away from lens then object have to be moved toward the lens .

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