5.What was the focal distance ( f ) when the radius of curvature was 0.7 and the

Question

5.What was the focal distance (f) when the radius of curvature was 0.7 and the index of refraction was 1.8?

6.Calculate the radius of curvature of a lens with a focal distance of 40 cm and an index of refraction of 1.2.

7. An object placed 35 cm away from a lens projects a real image equal to 55 cm behind the lens. What is the focal distance of the lens?

8. For the example above (7), what is the magnification of the lens?

9. An object 20 cm to the left of a convex lens is 1.0 m high. What is the height and location of its image if the lens has a magnification of -2.0?

__________________ m high and _________________ cm on the left / right side of the lens.

10. How did this simulation activity enhance your understanding of geometrical optics?

________________________________________________________________

Explanation / Answer

5)

for a convex lens

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

1/f = (1.8-1)*(2/0.7)

f = 0.4375 m

(6)

1/40 = (1.2-1)*(2/R)

R = 16 cm

(7)

from lens equation

1/s + 1/s' = 1/f

1/35 + 1/55 = 1/f

f = 21.4 cm

8)

m = s'/s = 55/35 = 1.57

(9)

m = hi/ho

-2 = hi/1

hi = 2 cm

-s'/s = m

-s'/20 = -2

s' = 40 cm

height of the image = 2 cm

0.02 m high and 40__ cm on the right side of the lens.

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