A lens is constructed of a plastic whose index of refraction is 1.5. the lens co

Question

A lens is constructed of a plastic whose index of refraction is 1.5. the lens convex on the left side and concave on the right side, with the radii of curvatures are 25 cm (convex) and 50 cm (concave) respectively.

a) Draw a diagram showing the lens
b) What will be the length of the focal length
c) Is the lens converging or diverging
d) refer to the lens of part (a) place an object 67 cm to the left of the lens and with the help of a NEW diagram:
i. calculate the position of the image produced by this object
ii. calculate the magnification of the image
iii. draw a principle ray diagram which helps locate this image
e) where should i place the object to obtain a real image at a distance of 300 cm to the right of the lens?
f) calculate this value and show it on your diagram.

Explanation / Answer

a)

b)
n = 1.5

R1 = +25cm
R2 = +50 cm

we know,

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

= (1.5-1)*(1/25 - 1/50)

= 0.5*0.02

f = +100 cm

c) f is positive. so, it is converging lense.

d)
f = 100
u = 67 cm (object distance)
v = ?(image distnace)
i)
1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/100 - 1/67

v = -203.03 cm ( image is same side as the object)

ii) m = -v/u = -(-203.03/67) = 3.03

iii)

e) m = +2
-v/u = 2

v = -2*u

1/u + 1/v = 1/f

1/u - 1/2*u = 1/f

1/2*u = 1/f

==> u = f/2 = 100/2 = 50 cm

so, the object should be placed 50 cm from the lense

f) u = 3m = 300 cm
1/u + 1/v = 1/f

1/v = 1/f - 1/u
1/v = 1/100 - 1/300

v = 150 cm (object distance)

g) for the second lense

object distance, u = 300-150 = 150 cm

1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/100 - 1/150

v = + 300 cm (final image distance from second lense)

the image is real and inverted.

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