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.
As shown in the diagram, when the incident ray is parallel to base of the prism, the emergent ray grazes along the second surface.
If refractive index of the material of prism is $\sqrt{2}$, the angle $\theta$ of prism is.