An object is placed in front of convex surface of a glass plano-convex lens of thickness `d = 9.0 cm`. The image of that object is formed on the plane surface of the lens serving as a screen. Find:
(a) the transverse magnification if the curvature radius of the lens's convex surface is `R = 2.5 cm`,
(b) the image illuminance if the luminance of the object i s`L = 7700 cd//m^(2)` and the entrance apertune diameter of the lens is `D = 5.0 mm`, losses of light are negligible.

Explanation

The formation of the image of a source `S`, placed at a distance `u` from the pole of the convex surface of plano-convex lens of thickness `d` is shown in the figure.
On applying the formula fro refraction through spherical surface, we get
`(n)/(s') - (1)/(s) = (n - 1)//R`, (here `n_(2) = n` and `n_(1) = 1`)
or, `(n)/(d) - (1)/(s) = (n - 1)//R` or, `(1)/(s) = (n)/(d) - ((n - 1))/(R)`
or, `(s')/(s) = s' {(n)/(d) - ((n - 1))/(R )}`
But in this case optical path of the light, corresponding to the distance `v` in the medium is `v//n`, so the magnification produced will be,
`beta = (s')/(ns) = (s')/(n) {(n)/(d) - ((n - 1))/(R )} = (d)/(n) {(n)/(d) - ((n - 1))/(R )} = 1 - (d(n - 1))/(nR)`
Subsituting the values, we get magnification `beta =- 0.20`.
(b) If the transverse area of the object is `A` (assumed small), the area o fthe image is `beta^(2) A`. we shell assume that `(pi D^(2))/(4) gt A`. Then light falling on the lens is : `LA (pi D^(2)//4)/(s^(2))`
from the definition of luminance (See Eqn. of the book , here `cos theta~~ 1` if `D^(2) lt lt s^(2)` and `d Omega = (pi D^(2)//4)/(s^(2)))`. Then the illuminance of the image is
`LA (pi D^(2) //4)/(s^(2)) // beta^(2) A = Ln^(2) pi D^(2)//4d^(2)`
Substitution gives `42 1x`.