4. Thin Lens A 1.00-cm-high object is placed 4.00 cm to the left of a converging

Question

4. Thin Lens A 1.00-cm-high object is placed 4.00 cm to the left of a converging lens of focal length 8.00 cm. A diverging lens of focal length -16.00 cm is 6.00 cm to the right of the converging lens. Find the position and height of the final image. Is the image inverted or upright? Real or virtual? 5. Spherical Mirror A spherical mirror is to be used to form an image 5.00 times the size of an object on a screen located 5.00 m from the object. (a) Is the mirror required concave or convex? (b) What is the required radius of curvature of the mirror? (c) Where should the mirror be positioned relative to the object?

Explanation / Answer

4)

from lens law

1/f = 1/v - 1/u

1/8 = 1/v + 1/4

v = - 8 cm

u for 2 lens = 6 + 8 = 14 cm

1/-16 = 1/v + 1/14

v = - 7.47 cm

final position is 7.47 cm front of the second lens.

b)

m1 = v/u = 8 /4 = 2

m2 = v/u = 7.47/14 = 0.533

H = 1 x m1 m2 = 1.067 cm

c)

virtual and upright   

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