Question

Derive the formula for the focal length of a thin lens.
OR
Explain the difference between diffraction and interference of light waves. Write down the formula for the fringe width in Young's double slit experiment. What will be the effect on the fringe width, when the separation between the slits is made one third and the distance between the slits and the screen is doubled? What will be the effect on the fringe width if the whole experiment is placed in water?

Hint:

For the lens: apply the single spherical-surface refraction formula twice (air-glass then glass-air) and add to get \( 1/f=(n-1)(1/R_1-1/R_2) \). For YDSE: \( \beta=\lambda D/d \); with \( d\to d/3,\ D\to 2D \) the width becomes \( 6\beta \), and in water it becomes \( \beta/n_w \).

Answer 1

Option 1: Lens maker's (focal length) formula

Step 1: Setup.
Consider a thin lens of material refractive index \( n \) (with respect to the surrounding medium) bounded by two spherical surfaces of radii of curvature \( R_1 \) and \( R_2 \). Let an object O lie on the principal axis. We use the refraction-at-a-single-spherical-surface formula twice.

Step 2: Refraction at the first surface.
For refraction from medium 1 (\( n_1=1 \)) into the lens (\( n_2=n \)) at a surface of radius \( R_1 \), the image would form at \( v_1 \):
\[ \frac{n}{v_1} - \frac{1}{u} = \frac{n-1}{R_1} \quad\text{...(1)} \]

Step 3: Refraction at the second surface.
The image \( I_1 \) at \( v_1 \) acts as the object for the second surface (radius \( R_2 \)), where light goes from the lens (\( n \)) back to the outside medium (\( 1 \)). For a thin lens the thickness is neglected, so the object distance is \( v_1 \):
\[ \frac{1}{v} - \frac{n}{v_1} = \frac{1-n}{R_2} \quad\text{...(2)} \]

Step 4: Add equations (1) and (2).
The \( n/v_1 \) terms cancel:
\[ \frac{1}{v} - \frac{1}{u} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]

Step 5: Introduce focal length.
By definition, when the object is at infinity (\( u\to\infty \)) the image forms at the focus, \( v=f \). Also for any object \( \dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f} \). Hence:
\[ \frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
This is the lens maker's formula giving the focal length of a thin lens.
\[\boxed{\dfrac{1}{f} = (n-1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)}\]

Option 2: Diffraction vs interference and fringe width

Step 1: Difference between interference and diffraction.
(i) Interference is the superposition of light from two (or more) separate coherent sources; diffraction is the superposition of the secondary wavelets coming from different parts of the same wavefront (a single slit/obstacle).
(ii) In interference all bright fringes are of equal width and equal intensity; in diffraction the central maximum is the brightest and widest, and the intensity of side maxima falls off rapidly.
(iii) In interference the dark fringes are almost perfectly dark; in diffraction the minima are not perfectly dark.

Step 2: Fringe width formula (YDSE).
\[ \beta = \frac{\lambda D}{d} \]
where \( \lambda \) is the wavelength, \( D \) the slit-to-screen distance and \( d \) the separation between the slits.

Step 3: Change the geometry.
New separation \( d' = \dfrac{d}{3} \) and new distance \( D' = 2D \). New fringe width:
\[ \beta' = \frac{\lambda D'}{d'} = \frac{\lambda (2D)}{d/3} = 6\,\frac{\lambda D}{d} = 6\beta \]
So the fringe width becomes six times the original.

Step 4: Immerse the apparatus in water.
In a medium of refractive index \( n_w \) the wavelength becomes \( \lambda' = \lambda/n_w \), so:
\[ \beta'' = \frac{\lambda' D}{d} = \frac{\beta}{n_w} \]
With \( n_w = 1.33 \), the fringe width reduces to about \( \beta/1.33 \approx 0.75\beta \), i.e. the fringes come closer together.
\[\boxed{\beta'=6\beta;\qquad \beta_{water}=\beta/n_w\approx 0.75\beta}\]