Physics M.C.Q (1 Marks)

Refraction is a phenomenon in which when a ray passes from one medium to another it bends away from its straight$-$line path due to the difference in optical densities or refractive indices of the two mediums. This bending is known as deviation.

As a glass has refractive index of $\frac{3}{2}$ with respect to air, the convergent light will bend towards the normal in the glass and again bend away from the normal, but in the whole process the beam is displaced parallel to initial path as shown in figure and converge at a greater distance from the glass slab.

Here initially $A\ \&\ B$ is parallel to each other after reflection by teh plane mirror $A'\ \&\ B'$ goes Parallel to each other.

Lateral displacement due to a slab is given by $\delta=\text{t}(\mu−1)$; where t is thickness of the medium and $\mu $ is refractive index of medium.

In combination $($refractive angles of prisms reversed with respect to each other$)$, the deviations through two prisms cancel out each other and the net deviation is due to the third prism only.

If $\mu$ is the refractive index and A is the angle of prism, then the angular dispersion produced by the prism will be given by $\delta=(\mu-1)\text{A}.$
Because the relative refractive index of glass with respect to water is small compared to the refractive of glass with respect to air, the dispersive power of the glass prism is more in air than that in water.

Before cut
$\frac{1}{\text{f}}=(\mu-1)\Big(\frac{2}{\text{R}}\Big)=4\text{D}$
After cut
$\frac{1}{\text{f}_1}=(\mu-1)\Big(\frac{1}{\text{R}}\Big)=\text{P}_1$
$\&\frac{1}{\text{f}_2}=(\mu-1)\Big(\frac{1}{\text{R}}\Big)=\text{P}_2$
Power of a divided lens will be = $P_1 + P_2$
$=(\mu-1)\Big(\frac{2}{\text{R}}\Big)$
$=4\text{D}$

Magnifaction of a mirror is defined as
$\text{m}=\frac{\text{size of image​}}{\text{size of object}}$
and since, in case of virtual and errect image, size of image and object both are positive. Hence magnification created by mirror is positive.

$\text{v}=(40-4)$
$\frac{1}{\text{f}}=\frac{1}{40-4}-\frac{1}{(-\text{u})}$
$\frac{\text{df}}{\text{du}}=0$ for f minimum.
$\frac{\text{df}}{\text{du}}=1-\frac{\text{u}}{20}=0$
$\text{u}=20$
$\text{f}_{\text{min}}=10\text{cm}$

If $\mu$ is the average refractive index and A is the angle of prism, then the angular dispersion produced by the prism is given by $\delta=(\mu-1)\text{A}.$

As velocity of wave is given by the relation $\text{v}=\text{f}\lambda$. When light ray goes from one medium to other medium, the frequency of light remains unchanged. Hence $\text{v}\propto\lambda$ or greater the wavelength, greater the speed.
The light of red colour is of highest wavelength and therefore of highest speed. Therefore, after travelling through the slab, the red colour emerges first.

Principal section is also defined as the normal 'side view' of the mirror for a ray diagram. In the diagram $AB$ is the principal section.

The ray $PQ$ of light passes through focus $F$ and incident on the concave mirror, after reflection, should become parallel to the principal axis and shown by ray $2$ in the figure.
Important points: We can locate the image of any extended object graphically by drawing any two of the following four special rays:
A ray initially parallel to the principal axis is reflected through the focus of the mirror $(1).$
A ray passing through the center of curvature is reflected back along itself $(3).$
A ray initially passing through the focus is reflected parallel to the principal axis $(2).$
A ray incident at the pole is reflected symmetrically.

Shift by which the object appears to be raised depends on these $4$ factors. The shift increases with increase in refraction index of medium and also with the thickness of denser medium.

The apparent vertical shift is given by the equation $\text{AD}=\frac{\text{RD}}{\mu}$, that is, ​$\mu=\frac{\text{RD}}{\text{AD}}$, where $AD$ is the apparent vertical shift, $RD$ is the real depth of the tank and $n$ is the refractive index of the denser medium, that is, water.
As real depth and refractive index is going to be constant always, the apparent shift or depth will be independent of viewing angle.
From the above equation, we can see that the apparent vertical shift is inversely proportional to the refractive index. That is, $\frac{1}{\mu}$​.

Here $P, P_1\ \&\ P_2$ are the power of Lenses.
$P = P_1+P_2$
$\frac{1}{\text{f}}=\frac{1}{\text{f}_1}+\frac{1}{\text{f}_2}$
$=(\mu-1)\Big(\frac{2}{\text{R}}\Big)+(\mu'-1)\Big(\frac{-1}{\text{R}}\Big)$
$=\Big(\frac{3}{2}-1\Big)\Big(\frac{2}{\text{R}}\Big)-\Big(\frac{4}{3}-1\Big)\Big(\frac{1}{\text{R}}\Big)$
$\frac{1}{\text{f}}=\frac{1}{\text{R}}-\frac{1}{3\text{R}}$
$\frac{1}{\text{f}}=\frac{3-1}{3\text{R}}$
$\text{f}=\frac{3\text{R}}{2}$
focal length of combined is positive means it will behave like a canvergent lens.

When water is poured then it is due to refraction of light the coin becomes visible. When the light rays travel from the water medium $($denser$)$ to the air medium $($rarer$)$ it bends away from the normal due to the refraction. Therefore, an image of the coin is formed at a smaller depth causing it to be visible $($as shown in the figure$)$.

Since, $\mu_\text{V}​>\mu_\text{R}$​, the shift is more for he violet light than the red light in a given medium.

We know that, $\text{f}=\frac{\text{R}}{\mu-1}$
Substituting $\text{R}=20\text{cm},\mu=1.5,$ we get
$\text{f}=\frac{20}{1.5-1}=40\text{cm}$
Since, the focal length is greater than zero.
Therefore, lens act as a convex lens irrespective of the side on which the object lies.

$\text{refractive index} = \frac{\text{actual depth​}}{\text{apparent depth}}$