I need help with number 1, after that, I should be able to figure the rest of th

Question

I need help with number 1, after that, I should be able to figure the rest of them out.

Notes: Regardless of problem specification, all lens or mirror optics problem solutions require ray diagrams with at least three principal rays. Additionally, assume that all lenses arc thin unless otherwise specified. Provide a complete description of all optical parameters (object/image locations and heights, focal lengths, magnification, all image properties, and a ray diagram) in problems 1- 4 below using standard sign conventions.

Explanation / Answer

1) r = 35 cm

f = r/2 = 35/2= 17.5 cm

concave mirror so f = +17.5 cm

a) 60 cm

1/f = 1/u + 1/v

1/17.5 = 1/60 + 1/v

v = 24.705 cm

m = -v/u = - 24.705/60 = -0.41176

negative sign indicates image is real and inverted

m = hi/ho = hi/15

hi = 0.41176*15 = 6.176 mm

image is real,inverted and diminished

b) 20 cm

1/f = 1/u + 1/v

1/17.5 = 1/20 + 1/v

v = 140 cm

m = -v/u = - 140/20 = -7

negative sign indicates image is real and inverted

m = hi/ho = hi/15

hi = 7*15 = 105 mm

image is real,inverted and enlarged

c) 7 cm

1/f = 1/u + 1/v

1/17.5 = 1/7 + 1/v

v = -11.667 cm

m = -v/u = (--11.667 / 7) = +1.67

positive sign indicates image is virtual and upright

m = hi/ho = hi/15

hi = 1.67*15 = 25 mm

image is virtual erect and enlarged.

2) r = 35 cm

f = r/2 = 35/2= 17.5 cm

convex mirror so f = -17.5 cm

a) 60 cm

1/f = 1/u + 1/v

1/-17.5 = 1/60 + 1/v

v = -13.54 cm

m = -v/u = -( -13.54/60) = +0.225

positive sign indicates image is virtual and upright

m = hi/ho = hi/15

hi = 0.225*15 = 3.387 mm

image is virtual,erect and diminished

b) 20 cm

1/f = 1/u + 1/v

1/-17.5 = 1/20 + 1/v

v = 9.33 cm

m = -v/u = - (-9.333/20) = 0.467

positive sign indicates image is virtual and upright

m = hi/ho = hi/15

hi = 0.467*15 = 7 mm

image is virtual erect and diminshed

c) 7 cm

1/f = 1/u + 1/v

1/-17.5 = 1/7 + 1/v

v = -5 cm

m = -v/u = -(-5 / 7) = +0.714

positive sign indicates image is virtual and upright

m = hi/ho = hi/15

hi = 0.714*15 = 10.71 mm

image is virtual erect and diminished.

the same goes with convex and concave lens

convex lens focal length is positive and concave lens focal length is negative

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