A 5.0 cm arrow (pointing up) is 110 cm from the 26 cm focal length convex (conve

Question

A 5.0 cm arrow (pointing up) is 110 cm from the 26 cm focal length convex (converging) lens.

A) How far behind the lens should we place a detection screen to get a clear image?

B) How tall will the arrow be on our screen and what direction will it point?

C) If we place the arrow 16 cm from the lens, what would the magnification be? Is the image in parts A and B magnified?

D) If we instead had a concave (diverging) lens, with a focal length of 26 cm, where is the image and what is its magnification for arrow 110 cm away from the lens?

Please show/ explain work. Thanks!

Explanation / Answer

For the convex lens

f = 26 cm

(A)

1/v - 1/u = 1/f

1/v - 1/-110 = 1/26

v = 34.05 cm

far behind the lens should we place a detection screen to get a clear image = 34.05 cm

(B)

Magnification = v/u = 34.05/-110 = -0.31

tall = 0.31*5 = 1.55 cm

So it will be 1.55 cm tall and point Downwards.

(C)

1/v -1/u = 1f

1/v - 1/-16 = 1/26

v = -41.6 cm

Magnification = v/u = -41.6/-110 = 0.378 Image is diminished.

In part A and B also image is diminished.

(D)

For concave lens

f = -26 cm

1/v-1/u = 1/f

1/v - 1/-110 = 1/-26

v = -21.03 cm

Image is 21.03 cm right of lens.

Magnification = v/u = -21.03/-110 = 0.19

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