Question

Four modifications are suggested in the lens formula to include the effect of the thickness t of the lens. Which one is likely to be correct?

Options

• \[\frac{1}{\nu} - \frac{1}{u} = \frac{t}{uf}\]

• \[\frac{t}{\nu^2} - \frac{1}{u} = \frac{1}{f}\]

• \[\frac{1}{\nu - t} - \frac{1}{u + t} = \frac{1}{f}\]

• \[\frac{1}{\nu} - \frac{1}{u} + \frac{t}{uv} = \frac{t}{f}\]

Answer 1

(c) \[\frac{1}{v - t} - \frac{1}{u + t} = \frac{1}{f}\] 

The thickness of lens will cause a shift in the position of object and image from the ideal condition, which will be: 
\[u_{new} = u + t(1 + \frac{1}{\mu} ) \text{ and } v_{new} = v - t(1 + \frac{1}{\mu} )\]
\[\text{ or } u_{new} = u + t + \frac{t}{\mu} \text{ and } v_{new} = v - t + \frac{t}{\mu}\]
As \[\frac{t}{\mu}\] will be quite small, it can be ignored.
Thus, u_new = u + t and v_new = v − t
We get the new lens formula:  \[\frac{1}{v - t} - \frac{1}{u + t} = \frac{1}{f}\]