A thin lens has a convex surface with radius of curvature of 25.5cm and a concav

Question

A thin lens has a convex surface with radius of curvature of 25.5cm and a concave surface with radius of curvature of 47 cm.  The index of refraction of the glass is 1.5.

(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 30cm from the lens.  Where is the lens?

(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

a)
b)
R1 = +25.5 cm

R2 = +47 cm

n = 1.5

1/f = (n-1)*(1/R1 - 1/R2)

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

f = +111.5 cm

the lense is converging lense

c) s(oject distance) = 30 cm

s'(image distance) =

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

1/s' = 1/f - 1/s
1/s' = 1/111.5 - 1/30

s' = -41.05 cm
d) image is virtual
e)
R1 = -47 cm

R2 = -22.5 cm

n = 1.5

1/f = (n-1)*(1/R1 - 1/R2)

1/f = (1.5 - 1)*(-1/47 + 1/25.5)

f = +111.5 cm

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