Between the primary and secondary rainbows, there is a dark band known as Alexandar's dark band. This is because
A magnifying glass is used, as the object to be viewed can be brought closer to the eye than the normal near point. This results in
An astronomical refractive telescope has an objective of focal length 20 m and an eyepiece of focal length 2 cm .
Will the focal length of a lens for red light be more, same or less than that for blue light?
As the refractive index for red is less than that for blue, parallel beams of light incident on a lens will be bent more towards the axis for blue light compared to red.
In other words, $\propto_b>\propto_r$
By lens maker's formula,
$$\frac{1}{f}=\left(n_{21}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$
Therefore, $f_b< f_t$.
Thus, the focal length for blue light will be smaller than that for red.
The near vision of an average person is 25 cm . To view an object with an angular magnification of 10 , what should be the power of the microscope?
The least distance of distinct vision of an average person (i.e., $D$ ) is 25 cm , in order to view an object with magnification 10,
Here, $v=D=25 \mathrm{~cm}$ and $u=f$
But the magnification $m=v / u=D / f$
$$\begin{aligned} & m=\frac{D}{f} \\ \Rightarrow\quad & f=\frac{D}{m}=\frac{25}{10}=2.5=0.025 \mathrm{~m} \\ & P=\frac{1}{0.025}=40 \mathrm{D} \end{aligned}$$
This is the required power of lens.