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: Focal length of a thin lens

Step 1: Idea of the derivation.
A lens is just two refracting spherical surfaces placed back to back. If we track the image formed by the first surface and feed it as the object to the second surface, the two single-surface refraction relations add up to the lens formula. The single-surface relation used is \(\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}\).

Step 2: First surface (air to glass, radius \(R_1\)).
With \(n_1=1,\ n_2=n\), the intermediate image sits at \(v_1\):
\[ \frac{n}{v_1}-\frac{1}{u}=\frac{n-1}{R_1} \]

Step 3: Second surface (glass to air, radius \(R_2\)).
Now \(n_1=n,\ n_2=1\), and because the lens is thin the object distance is still \(v_1\):
\[ \frac{1}{v}-\frac{n}{v_1}=\frac{1-n}{R_2} \]

Step 4: Superpose.
Adding the two equations removes \(v_1\) and gives \(\dfrac{1}{v}-\dfrac{1}{u}=(n-1)\!\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)\).

Step 5: Identify \(1/f\).
Since \(\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\) by the thin-lens definition, we read off
\[\boxed{\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)}\]
which fixes the focal length from the material and the two curvatures.

Option 2: Diffraction, interference and fringe width

Step 1: Contrast in one line each.
Interference needs two independent coherent beams that overlap; diffraction arises from one wavefront bending round an edge or through a slit. Interference bands are uniform in width and brightness with truly dark gaps; diffraction bands shrink and dim away from a dominant central band and its minima are only partly dark.

Step 2: Standard fringe width.
For Young's experiment \(\beta=\dfrac{\lambda D}{d}\), linear in \(\lambda\) and \(D\) and inverse in slit spacing \(d\).

Step 3: Scale the two lengths.
Replace \(d\to d/3\) (denominator shrinks, \(\times3\)) and \(D\to2D\) (numerator grows, \(\times2\)). Net factor \(2\times3=6\), so \(\beta\to6\beta\): fringes get six times wider.

Step 4: Effect of water.
Inside water of index \(n_w\) light slows and its wavelength shrinks to \(\lambda/n_w\), so every fringe narrows by the same factor: \(\beta\to\beta/n_w\). Taking \(n_w\approx1.33\) gives roughly \(0.75\beta\), the pattern tightening up.
\[\boxed{\beta_{new}=6\beta,\qquad \beta_{water}=\beta/n_w\approx0.75\beta}\]