A double-convex lens is to be made of glass with an index of refraction of 1.5.One surface is to have twice the radius of curvature of the other and the focal length is to be 60mm. What is the (a) smaller and (b) larger radius?

Short Answer

Expert verified
  1. Smaller radius of the lens is 45mm
  2. Larger radius of the lens is 90mm

Step by step solution

01

Listing the given quantities

Refractive index of lens, n=1.5

Focal length,f=60.0mm

Radius of one surface is twice the other, i.e.,r2=-2r1

02

Understanding the concepts of lens maker equation

By using the Lens maker’s formula, we can find the radius of both curvatures of the lens.

Formula:

Lens maker’s formula

1f=(n-1)(1r1-1r2)...34-10

03

(a) Calculations of the smaller radius of the lens

Lens maker’s formula is given by equation,

1f=n-11r1-1r2

Since one surface has twice the radius of the other and since one surface is convex to the incoming light while the other is concave, we can write,

r2=-2r1

So, we can get,

1fn-1=1r1+12r1=32r1r1=32fn-1=1.560.01.5-1=45mm

Smaller radius of the lens is 45mm.

04

(b) Calculations of the larger radius of the lens

As per the condition given in the problem, r2will be double of r1

So, r2 = 90 mm

Larger radius of the lens is 90 mm

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Isaac Newton, having convinced himself (erroneously as it turned out) that chromatic aberration is an inherent property of refracting telescopes, invented the reflecting telescope, shown schematically in Fig. 34-59. He presented his second model of this telescope, with a magnifying power of 38, to the Royal Society (of London), which still has it. In Fig. 34-59, incident light falls, closely parallel to the telescope axis, on the objective mirror. After reflection from the small mirror (the figure is not to scale), the rays form a real, inverted image in the focal plane (the plane perpendicular to the line of sight, at focal point F). This image is then viewed through an eyepiece. (a) Show that the angular magnification for the device is given by Eq. 34-15:

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fob

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