Question 13 Marks
Find the angle of minimum deviation for an equilateral prism made of a material of refractive index 1.732. What is the angle of incidence for this deviation?
Answer
View full question & answer→Refractive index $(\mu)$ of the material from which prism is made = 1.732 We know refractive index is given by:$\mu=\frac{\Big[\frac{\delta_{\text{min}}+\text{A}}{2}\Big]}{\sin\Big[\frac{\text{A}}{2}\Big]}$
Where $\delta_{\text{min}}$ is the angleof minimum deviation and A is the angle of prism = 60º$\Rightarrow 1.732\times \sin(30^\circ)=\sin\Big(\frac{\delta_{\text{min}}+60^\circ}{2}\Big)$
$\Rightarrow \frac{1.732}{2}=\sin\Big(\frac{\delta_{\text{min}}+60^\circ}{2}\Big)$
$\Rightarrow \Big(\frac{\delta_{\text{min}}+60^\circ}{20}\Big)=60^\circ$
$\delta_{\text{min}}=60^\circ$
$\delta_{\text{min}}=2\text{i}-\text{A}$
$2\text{i}=120^\circ$
$\text{i}=60^\circ$
Hence, the required angle of deviation is 60°.
Where $\delta_{\text{min}}$ is the angleof minimum deviation and A is the angle of prism = 60º$\Rightarrow 1.732\times \sin(30^\circ)=\sin\Big(\frac{\delta_{\text{min}}+60^\circ}{2}\Big)$$\Rightarrow \frac{1.732}{2}=\sin\Big(\frac{\delta_{\text{min}}+60^\circ}{2}\Big)$
$\Rightarrow \Big(\frac{\delta_{\text{min}}+60^\circ}{20}\Big)=60^\circ$
$\delta_{\text{min}}=60^\circ$
$\delta_{\text{min}}=2\text{i}-\text{A}$
$2\text{i}=120^\circ$
$\text{i}=60^\circ$
Hence, the required angle of deviation is 60°.
The presence of air medium in between the sheets does not affect the shift. The shift will be due to 3 sheets of different refractive index other than air.$\Delta \text{t}=\Big[1-\frac{1}{\mu_1}\Big]\text{t}_1+\Big[1-\frac{1}{\mu_2}\Big]\text{t}_2+\Big[1-\frac{1}{\mu_3}\Big]\text{t}_3$
Given, P = 5 diopter (convex lens)$\Rightarrow\text{f}=\frac{1}{5}\text{m}=20\text{cm}$
Height of the lake = 2.5m
Applying snell's law,
Radius of the cylindrical glass tube = 1cm We know, $\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$ Here, $\text{u}=-8\text{cm}, \ \mu_2=\frac{3}{2},\mu_1=\frac{4}{3},\text{R}=+1\text{cm}$ So, $\frac{3}{2\text{v}}+\frac{4}{3\times8}\Rightarrow\frac{3}{2\text{v}}+\frac{1}{6}=\frac{1}{6}$$ \ \text{v}=\infty$
Thickness of glass $=3\text{cm}, \ \mu_{\text{g}}=1.5$ Image shif $=3\Big(1-\frac{1}{1.5}\Big)$ [Treating it as a simple refraction problem because the upper surface is flat and the spherical surface is in contact with the object]$=3\times\frac{0.5}{1.5}=1\text{cm}.$
Given AB = 3cm, u = -7.5cm, f = 6cm. Using $\Rightarrow\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}$ Putting values according to sign conventions,$\frac{1}{\text{v}}=\frac{1}{6}-\frac{1}{-7.5}=\frac{3}{10}$
u = -30cm, R = -40cm
Given, $\mu=1.5$ And angle of prism $=4^{\circ}$$\therefore \ \mu=\frac{\sin\big(\frac{\text{A}+\delta_{\text{m}}}{2}\big)}{\frac{\sin\text{A}}{2}}=\frac{\big(\frac{\text{A}+\delta_{\text{m}}}{2}\big)}{\big(\frac{\text{A}}{2}\big)}$ $\big($for small angle $\sin\theta=\theta\big)$
Let the object be placed at a height x above the surface of the water. The apparent position of the object with respect to mirror should be at the centre of curvature so that the image is formed at the same position. Since, $\frac{\text{Real depth}}{\text{Apparent depth}}=\frac{1}{\mu}$ (with respect to mirror) Now, $\frac{\text{x}}{\text{R}-\text{h}}=\frac{1}{\mu}$$\Rightarrow\text{x}=\frac{\text{R}-\text{h}}{\mu}$
$\text{u}=-30\text{cm}, \ \text{f}=-20\text{cm}$
Shadow length $= \text{BA}' = \text{BD} + \text{A}'\text{D} = 0.5 + 0.5 \tan \text{r}$
For the lens, f = 15cm, u = 30cm From lens formula $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$$\Rightarrow\frac{1}{\text{v}}=\frac{1}{15}-\frac{1}{30}=\frac{1}{30}\Rightarrow\text{v}=30\text{cm}$
$\mu_1=1, \ \mu_2=1.5,$
If the image in the mirror will form at the focus of the converging lens, then after transmission through the lens the rays of light will go parallel. Let the object is at a distance x cm from the mirror$\therefore$ u = -x cm; v = 25 - 15 = 10cm (because focal length of lens = 25cm)
Height of the object AB = 1.6cm Diameter of the ball bearing = d = 0.4cm ⇒ R = 0.2cm Given, u = 20cm We know, $\frac{1}{\text{u}}+\frac{1}{\text{v}}=\frac{2}{\text{R}}$ Putting the values according to sign conventions $\frac{1}{-20}+\frac{1}{\text{v}}=\frac{2}{0.2}$$\Rightarrow\frac{1}{\text{v}}=\frac{1}{20}+10=\frac{201}{20}\Rightarrow\text{v}=0.1\text{cm}=1\text{mm}$ inside the ball bearing.