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 n1, to another of refractive index n2.
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 begin mathsize 12px style angle end styleNOM = MN / OM ,      tan begin mathsize 12px style angle end styleNCM = MN / MC ,    tan begin mathsize 12px style angle end styleNIM = 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    n1 × sin i = n2 × sin r or for small angles   n1 × i = n2 × r
 
Substituting i and r from Eqs. (1) and (2), we get,   begin mathsize 12px style fraction numerator n subscript 1 over denominator O M end fraction space plus space fraction numerator n subscript 2 over denominator M I end fraction space equals space fraction numerator n subscript 2 space minus space n subscript 1 over denominator M C end fraction end style .............................................(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         begin mathsize 12px style n subscript 2 over v space minus space n subscript 1 over u space equals space fraction numerator n subscript 2 minus n subscript 1 over denominator R end fraction end style .............................................................(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 I1 of the object O [ middle figure]. The image I1 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
 
begin mathsize 12px style fraction numerator n subscript 1 over denominator O B end fraction space plus space fraction numerator n subscript 2 over denominator B I subscript 1 end fraction space equals space fraction numerator n subscript 2 space minus space n subscript 1 over denominator B C subscript 1 end fraction end style  .............................(5)
A similar procedure applied to the second interface  ADC gives,
 
begin mathsize 12px style negative fraction numerator n subscript 2 over denominator D I subscript 1 end fraction space plus space fraction numerator n subscript 1 over denominator D I end fraction space equals space fraction numerator n subscript 2 space minus space n subscript 1 over denominator D C subscript 2 end fraction end style  ..............................(6)
Note that now the refractive index of the medium on the right side of ADC is n1, while on its left it is n2.
Further DI1 is negative as the distance is measured against the direction of incident light.
 
For a thin lens, BI1 = DI1. Adding Eqs. (5) and (6), we get
 
begin mathsize 12px style fraction numerator n subscript 1 over denominator O B end fraction plus fraction numerator n subscript 1 over denominator D I end fraction space equals space left parenthesis n subscript 2 minus n subscript 1 right parenthesis open parentheses fraction numerator 1 over denominator B C subscript 1 end fraction space plus space fraction numerator 1 over denominator D C subscript 2 end fraction close parentheses end style ..........................(7)

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

 begin mathsize 12px style n subscript 1 over f space equals space left parenthesis n subscript 2 minus n subscript 1 right parenthesis open parentheses fraction numerator 1 over denominator B C subscript 1 end fraction space plus space fraction numerator 1 over denominator D C subscript 2 end fraction close parentheses end style .......................................(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,
 
BC1 = + R1DC2 = –R2   and we have begin mathsize 12px style space space n subscript 21 space equals space n subscript 2 over n subscript 1 end style
  So Eq. (8) can be written as  begin mathsize 12px style 1 over f space equals space open parentheses n subscript 21 space minus space 1 close parentheses space open parentheses 1 over R subscript 1 space minus space 1 over R subscript 2 close parentheses space space end style  .......................(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 R1is negative, R2 positive and
therefore, f is negative.
 
 

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

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