Note: It would be great to have someone answer with a picture where they draw an

Question

Note: It would be great to have someone answer with a picture where they draw and specific how to get the answer. Thanks

A thin lens has a convex surface whose radius of curvature is 25.5 cm and a concave surface whose radius of curvature is 47.0 cm. The index of refraction of the glass is 1.50. (a) Draw a sketch of the lens. (b)What is the focal length of the lens? Is the lens converging or diverging? (c) A real point object is located 30 cm from the lens. Where is the image? (d) Is the image real or virtual? (e) If you reverse the lens so that the concave surface is the first surface, show that the lensmaker's equation gives the same result for the focal length.

Explanation / Answer

1/F = (u-1) *((1/R1)-(1/R20

1/F = (1.5-1)*((1/25.5)-(1/47))

F = + 111.488cm

converging lens

c) 1/S +1/S' = 1/F

1/30 + 1/S' = 1/111.488

S'= ?41.04 cm

d) virtual

e) 1/F = (1.5-1)*((-1/47)-(-1/25.5))

F = 111.488 cm

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