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Physics 5 Marks Questions

Dispersion and Spectra · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 3 questions.

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Question 15 Marks
Three thin prisms are combined as shown in figure. The refractive indices of the crown glass for red, yellow and violet rays are $\mu_\text{r},\mu_\text{y}$ and $\mu_\text{v}$ respectively and those for the flint glass are $\mu'_\text{r},\mu'_\text{y}$ and $\mu'_\text{v}$ respectively. Find the ratio $\frac{\text{A}'}{\text{A}}$ for which.

  1. There is no net angular dispersion.
  2. There is no net deviation in the yellow ray.
Answer
Total deviation for yellow ray produced by the prism combination is,

$\delta_\text{y}=\delta_\text{cy}-\delta_\text{fy}+\delta_\text{cy}=2\delta_\text{cy}-\delta_\text{fy}$ $=2(\mu_\text{cy}-1)\text{A}-(\mu_\text{cy}-1)\text{A}'$

Similarly the angular dispersion produced by the combination is

$\delta_\text{v}-\delta_\text{r}=\big[(\mu_\text{vc}-1)\text{A}-(\mu_\text{vf}-1)\text{A}'+(\mu_\text{vc}-1)\text{A}\big]\\\big[(\mu_\text{rc}-1)\text{A}-(\mu_\text{rf}-1)\text{A}'+(\mu_\text{r}-1)\text{A})\big]$

$=2(\mu_\text{vc}-1)\text{A}-(\mu_\text{vf}-1)\text{A}'$

  1. For net angular dispersion to be zero,

$\delta_\text{v}-\delta_\text{r}=0$

$\Rightarrow2(\mu_\text{vc}-1)\text{A}=(\mu_\text{vf}-1)\text{A}'$

$\Rightarrow\frac{\text{A}'}{\text{A}}=\frac{2(\mu_\text{cv}-\mu_\text{rc})}{(\mu_\text{vf}-\mu_\text{rf}}$

$=\frac{2(\mu_\text{v}-\mu_\text{r})}{(\mu'_\text{v}-\mu'_\text{r})}$

  1. For net deviation in the yellow ray to be zero,

$\delta_\text{y}=0$

$\Rightarrow2(\mu_\text{cy}-1)\text{A}=(\mu_\text{fy}-1)\text{A}'$

$\Rightarrow\frac{\text{A}'}{\text{A}}=\frac{2(\mu_\text{cy}-1)}{(\mu_\text{fy}-1)}$

$=\frac{2(\mu_\text{y}-1)}{(\mu'_\text{y}-1)}$

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Question 25 Marks
A thin prism of angle $6.0^\circ,\omega=0.07$ and $\mu_\text{y}=1.50$ is combined with another thin prism having $\omega=0.08$ and $\mu_\text{y}=1.60.$ The combination produces no deviation in the mean ray.
  1. Find the angle of the second prism.
  2. Find the net angular dispersion produced by the combination when a beam of white light passes through it.
  3. If the prisms are similarly directed, what will be the deviation in the mean ray
  4. Find the angular dispersion in the situation described in (c).
Answer
Given that, $\text{A}'=6^\circ,\ \omega'=0.07,\ \mu'_\text{y}=1.50$

$\text{A}=?\ \omega=0.08,\ \mu_\text{y}=1.60$

The combination produces no deviation in the mean ray.

  1. $\delta_\text{y}=(\mu_\text{y}-1)\text{A}-(\mu'_\text{y}-1)\text{A}'=0$ [Prism must be oppositely directed]

$\Rightarrow(1.60-1)\text{A}=(1.50-1)\text{A}'$

$\Rightarrow\text{A}=\frac{0.50\times6^\circ}{0.60}=5^\circ$

  1. When a beam of white light passes through it,

Net angular dispersion $=(\mu_\text{y}-1)\omega\text{A}-(\mu'_\text{y}-1)\omega'\text{A}'$

$\Rightarrow(1.60-1)(0.08)(5^\circ)-(1.50-1)(0.07)(6^\circ)$

$\Rightarrow0.24^\circ-0.21^\circ=0.03^\circ$

  1. If the prisms are similarly directed,

$\delta_\text{y}=(\mu_\text{y}-1)\text{A}+(\mu'_\text{y}-1)\text{A}$

$=(1.60-1)5^\circ+(1.50-1)6^\circ=3^\circ+3^\circ=6^\circ$

  1. Similarly, if the prisms are similarly directed, the net angular dispersion is given by,

$\delta_\text{v}-\delta_\text{r}=(\mu_\text{y}-1)\omega\text{A}-(\mu'\text{y}-1)\omega'\text{A}'$ $=0.24^\circ+0.21^\circ=0.45^\circ$

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Question 35 Marks
The refractive index of a material M1 changes by 0.014 and that of another material M2 changes by 0.024 as the colour of the light is changed from red to violet. Two thin prisms one made of M1(A = 5.3°) and other made of M2(A = 3.7°) are combined with their refracting angles oppositely directed.
  1. Find the angular dispersion produced by the combination.
  2. The prisms are now combined with their refracting angles similarly directed. Find the angular dispersion produced by the combination.
Answer
Given that, $\mu'_\text{v}-\mu'_\text{r}=0.014$ and $\mu_\text{v}-\mu_\text{r}=0.024$ $\text{A}' = 5.3^\circ\ \text{and A} = 3.7^\circ$

  1. When the prisms are oppositely directed,

angular dispersion $=(\mu_\text{v}-\mu_\text{r})\text{A}-(\mu'_\text{v}-\mu'_\text{r})\text{A}'$

$=0.024\times3.7^\circ-0.014\times5.3^\circ=0.0146^\circ$

  1. When they are similarly directed,

angular dispersion $=(\mu_\text{v}-\mu_\text{r})\text{A}+(\mu'_\text{v}-\mu'_\text{r})\text{A}'$

$=0.024\times3.7^\circ+0.014\times5.3^\circ=0.163^\circ$

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