Establish the formula of lens.

1/f=(meu-1)(1/R-1/R')

### Asked by Ravishankarpoddar652 | 3rd Jan, 2019, 09:22: AM

Expert Answer:

###
Figure 1 shows the geometry of formation of image I of an object O on the principal axis of a spherical surface with centre of curvature C,
and radius of curvature R. The rays are incident from a medium of refractive index n_{1}, to another of refractive index n_{2}.
We take the aperture (or the lateral size) of the surface to be small compared to other distances involved, so that small angle approximation
can be made. In particular, NM will be taken to be nearly equal to the length of the perpendicular from the point N on the principal axis.
We have, for small angles,

tan NOM = MN / OM , tan NCM = MN / MC , tan NIM = MN / MI
Now, for ΔNOC, i is the exterior angle. Therefore, i = ∠NOM + ∠NCM

i = (MN/OM) + (MN/MC) ......................................................................................................(1)

Similarly, r = ∠NCM – ∠NIM, i.e., r = (MN/MC) - (MN/MI) ...............................................(2)

Now, by Snell’s law n_{1} × sin i = n_{2} × sin r or for small angles n_{1} × i = n_{2} × r
Substituting i and r from Eqs. (1) and (2), we get, .............................................(3)

Here, OM, MI and MC represent magnitudes of distances. Applying the Cartesian sign convention,

OM = –u, MI = +v, MC = +R

Substituting these in Eq. ( 3 ), we get .............................................................(4)

Equation ( 4 ) gives us a relation between object and image distance in terms of refractive index of the medium and the radius of

curvature of the curved spherical surface. It holds for any curved spherical surface.
Left most Figure above 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 I_{1} of the object O [ middle figure]. The image I_{1} acts as a virtual object for the second surface

that forms the image at I [ Right most figure]. Applying Eq. (4) to the first interface ABC, we get
.............................(5)
A similar procedure applied to the second interface ADC gives,
..............................(6)
Note that now the refractive index of the medium on the right side of ADC is n_{1}, while on its left it is n_{2}.
Further DI_{1} is negative as the distance is measured against the direction of incident light.
For a thin lens, BI_{1} = DI_{1}. Adding Eqs. (5) and (6), we get
..........................(7)

Suppose the object is at infinity, i.e., OB → ∞ and DI = f, Eq. (7) gives

.......................................(8)

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. A lens has two foci, F and F′, on either side of it . By the sign convention,

BC_{1} = + R_{1}, DC_{2} = –R_{2} and we have
So Eq. (8) can be written as .......................(9)

Equation (9) 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 R_{1}is negative, R_{2} positive and

therefore, f is negative.

_{1}, to another of refractive index n

_{2}.

tan NOM = MN / OM , tan NCM = MN / MC , tan NIM = MN / MI

i = (MN/OM) + (MN/MC) ......................................................................................................(1)

_{1}× sin i = n

_{2}× sin r or for small angles n

_{1}× i = n

_{2}× r

Here, OM, MI and MC represent magnitudes of distances. Applying the Cartesian sign convention,

Substituting these in Eq. ( 3 ), we get .............................................................(4)

Equation ( 4 ) gives us a relation between object and image distance in terms of refractive index of the medium and the radius of

curvature of the curved spherical surface. It holds for any curved spherical surface.

(i) The first refracting surface forms the image I

_{1}of the object O [ middle figure]. The image I

_{1}acts as a virtual object for the second surface

that forms the image at I [ Right most figure]. Applying Eq. (4) to the first interface ABC, we get

_{1}, while on its left it is n

_{2}.

_{1}is negative as the distance is measured against the direction of incident light.

_{1}= DI

_{1}. Adding Eqs. (5) and (6), we get

Suppose the object is at infinity, i.e., OB → ∞ and DI = f, Eq. (7) gives

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. A lens has two foci, F and F′, on either side of it . By the sign convention,

_{1}= + R

_{1}, DC

_{2}= –R

_{2}and we have

Equation (9) 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 R

_{1}is negative, R

_{2}positive and

therefore, f is negative.

### Answered by Thiyagarajan K | 3rd Jan, 2019, 04:43: PM

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