THESE 2 QUESTIONS GO TOGETHER IN ORDER TO SOLVE 5) A meniscus lens is to be made
ID: 1412439 • Letter: T
Question
THESE 2 QUESTIONS GO TOGETHER IN ORDER TO SOLVE
5) A meniscus lens is to be made of plastic with n = 1.80. The concave surface is to have a radius of curvature of 13.0 cm. Find the radius of curvature of the convex surface if the lens is to be:
(a) a diverging lens with a focal length of -40.0 cm, or (b) a converging lens with a focal length of +54.0 cm.
6) The diverging lens in Part (a) of Problem 5 is to be used in a pair of glasses. It in the shape of a circle with a diameter of 7.50 cm, and is 3.71 mm thick at the center, how thick is the edge of the lens? I AM STUCK DO NOT KNOW HOW TO GET 6 the answer is 6mm BUT I DO NOT KNOW HOW TO GET IT! thanks!
Explanation / Answer
Hi,
5) Assuming that the equation for thin lens is valid in this case, then we have the following:
1/f = (n - 1) (1/R1 - 1/R2) ; where f is the focal distance, n is the index of refraction and R1 and R2 are the radii of curvature of each surface of the lens.
(a) Using the previous equation and considering that f = -40 cm ; R1 = 13 cm and n = 1.8 then:
1/R2 = 1/R1 - 1/[f*(n-1)] = 1/13 - 1/[-40*(1.8 - 1)] = 0.108 ::::::::::: R2 = 9.2 cm
(b) Proceeding in a similar way as the question before, but with f = 54 cm ; R1 = 13 cm and n = 1.8 :
1/R2 = 1/13 - 1/[54*(1.8 - 1)] = 0.054 ::::::::: R2 = 18.5 cm
6) In this case we can use some additional equations to find the answer:
s = R - (R2 - y2)1/2 ; where S is the sag of the lens, R is the radius of curvature and y is half the diameter of the glass.
e = t - s1 + s2 ; where e is the thickness of the edge, t is the thickness of the center and s1 and s2 are the sags of the lenses.
So:
s1 = 13 - (132 - 3.752)1/2 = 0.55 cm
s2 = 9.2 - (9.22 - 3.752)1/2 = 0.80 cm
e = 3.71 mm - 5.5 mm + 8 mm = 6.21 mm
Note: the answer is not exactly 6 mm, which could be because the supposition of thin lens at problem 5 was not correct or because during calculations some approximations were made.
I hope it helps.
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