1) A thin concave-convex lens has its concave surface of radius 20cm and its con

Question

1) A thin concave-convex lens has its concave surface of radius 20cm and its convex surface of radius 10cm. Its index of refraction is 1.5, and an object of height 5cm is located 50 cm in front of the concave side of the lens along its principle axis. A)What is the focal length of the lens? Is it converging or diverging? B) Where is the image of the light bulb located? Is the image real or virtuall. C) What is the height of the image? Is the image enlarged or reduced? Erect or inverted?

2) A reflecting telescope hasa a primary mirror of diameter 20cm and radius of curvature 2m. If the telescope is pointed at the moon, 3480 Km in diameter and 386,000km away, Find A)The location of the moon's image. B) The size of the moon's image. C) Describe the moon's image: Real/virtual, erect/inverted, enlarged/reduced, and normal/perverted

Explanation / Answer

1)Given,

R1 = 20 cm and R2 = 10 cm

n = 1.5 ; h = 5 cm ; o = 50 cm

If R2 = R than R1 = 2R (R = 10)

(a)we know for the system of two lenses:

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

1/f = (n - 1) [1/2R - 1/R]

1/f = - (n - 1) / 2 R

f = -2R/(n-1)

f = - 2 x 10 / (1.5 - 1) = -40 cm

Hence, f = -40 cm.

Since the focal length is negative, its a diverging or concave lens.

(b)we have, object distance = o = 50 cm and focal length = -40 cm

Let i be the image distance. we know from lens equation that

1/f = 1/i + 1/o

i = o x f / (o - f) = 50 x -40 / (50 + 40) = -22.22 cm

Hence, image distance = i = -22 cm.

c)Let h' be the image height.

we know that, magnification is

M = -i/o = h'/h

-(-22.22)/50 = h'/5

h' = 22.22 x 5 / 50 = 2.222

Hence, the image height = h' = 2.22 cm.

The image is reduced in size but erected.

[Pls put the other question seperately... thanks. Let me know for any query].

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