This problem involves calculating the focal length of the second lens in a two-lens system designed to expand a collimated light beam. The system configuration is a Keplerian beam expander, where the input and output beams are collimated.
The magnification ($M$) of a Keplerian beam expander is the ratio of the final beam diameter to the initial beam diameter:
$ M = \frac{d_2}{d_1} $
Substituting the given values:
$ M = \frac{14 \text{ mm}}{2 \text{ mm}} = 7 $
For a Keplerian beam expander, the magnification is also given by the ratio of the focal lengths of the two lenses:
$ M = \frac{f_2}{f_1} $
To find $f_2$, we rearrange the formula:
$ f_2 = M \times f_1 $
Substituting the calculated magnification and the given $f_1$:
$ f_2 = 7 \times 40 \text{ mm} $
$ f_2 = 280 \text{ mm} $
The required focal length for the second lens is $280 \text{ mm}$. This value falls within the specified range.
A ring is placed at the bottom of a trough containing four immiscible liquids of refractive indices $1.0, 2.0, 3.0$ and $4.0$ poured one above the other of heights $10$ cm, $20$ cm, $30$ cm, and $40$ cm respectively. The apparent depth at which the ring appears, when seen from outside is____________ cm.
A plane polarized light is incident on a polariser with its pass axis making angle $\theta$ with x-axis, as shown in the figure. At four different values of $\theta$, $\theta = 8^\circ, 38^\circ, 188^\circ$ and $218^\circ$, the observed intensities are same. What is the angle between the direction of polarization and x-axis ?
