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Use the Above Relation to Obtain the Condition on the Position of the Object and the Radius of Curvature in Terms of N1and N2 When the Real Image is Formed. - Physics

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Question

Use the above relation to obtain the condition on the position of the object and the radius of curvature in terms of n1and n2 when the real image is formed.

Answer in Brief
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Solution

(a) The position of the object, and the image formed by a double convex lens,

(b) Refraction at the first spherical surface and (c) Refraction at the second spherical surface.

Figure (a) shows the geometry of image formation by a double convex lens. The image formation can be seen in terms of two steps: (i) The first refracting surface forms the image I1 of the object O [Fig. (b)]. The image I1 acts as a virtual object for the second surface that forms the image at I [Fig. (c)].

Applying Equation `"n"_2/"v"-"n"_1/"u" = ("n"_2-"n"_1)/"R"` to the first interface ABC, 

We get
`"n"_1/"OB" + "n"_2/"BI"_1 = ("n"_2-"n"_1)/"BC"_1` ......(1)

A similar procedure applied to the second interface ADC gives,

`"n"_2/"DI"_1 +"n"_1/"DI" = ("n"_2-"n"_1)/"DC"_1`   ......(2)

For a thin lens, BI1= DI1. Adding
Eqs. (1) and (2), we get

`"n"_1/"OB" + "n"_1/"DI" = ("n"_2 - "n"_1)(1/"BC"_1+1/"DC"_2)`.....(3)

Suppose the object is at infinity, i.e., `"OB" -> ∞  and  "DI" = f, Eq.(3) "gives"`

`"n"_1/f = ("n"_2-"n"_2)(1/"BC"_1 + 1/"DC"_2) ......(4)`

The point where image of an object placed at infinity is formed is called the focus F, of the lens and the distance f gives its focal length.

By the sign convention,
`"BC"_1 = + "R"_1`
`"DC"_2 = -"R"_2`

So Eq.(4) can be written as
`1/f = ("n"_21 -1)(1/"R"_1 - 1/"R"_2).........(5)  (∵ "n"_21 = "n"_2/"n"_1)`

Equation (5) is known as the lens maker’s formula. It is useful to design lenses of desired focal length using surfaces of suitable radii of curvature. Note that the formula is true for a concave lens also. In that case R1is negative, R2is positive and therefore, f is negative.

From Eqs. (3) and (4), we get
`"n"_1/"OB" + "n"_1/"DI" = "n"_1/f`...............(6)

Again, in the thin lens approximation, B and D are both close to the optical centre of the lens. Applying the sign convention,

BO = –u, DI = +v,we get

`1/"v"-1/"u"=1/f`.............(7)

Equation (7) is the familiar thin lens formula.

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2014-2015 (March) Ajmer Set 2

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