Physics 4 Marks Question

1. As shown in the figure, $\sin\text{i}=\frac{15}{25}$

So, $\frac{\sin\text{i}}{\sin\text{r}}=\frac{1}{\mu}=\frac{3}{4}$

$\Rightarrow\sin\text{r}=\frac{4}{5}$

Again, $\frac{\text{x}}{2}=\tan\text{r}$ (from figure)

So, $\sin\text{r}=\frac{\tan\text{r}}{\sqrt{1+\tan^2\text{r}}}=\frac{\frac{\text{x}}{2}}{\sqrt{1+\frac{\text{x}^2}{4}}}$

$\Rightarrow\frac{\text{x}}{\sqrt{{4}+\text{x}^2}}=\frac{4}{5}$

$\Rightarrow25\text{x}^2=16(4+\text{x}^2)\Rightarrow9\text{x}^2=64\Rightarrow\text{x}=\frac{8}{3}\text{m}$

$\therefore$ Total radius of shadow $=\frac{8}{3}+0.15=2.81\text{m}$

2. For maximum size of the ring, i = critical angle = C

Let, R = maximum radius

$\Rightarrow\sin\theta_\text{C}=\frac{\sin\theta_\text{C}}{\sin\text{r}}=\frac{\sin\text{R}}{\sqrt{20^2+\sin\text{R}^2}}=\frac{3}{4}$(since, sin r = 1)

$\Rightarrow16\text{R}^2=9\text{R}^2+9\times400$

$\Rightarrow7\text{R}^2=9\times400$

$\Rightarrow\text{R}=22.67\text{cm.}$

1. Image seen from left:

u = (5 - 15) = -3.5cm

R = -5cm

$\therefore \ \frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$

$\Rightarrow\frac{1}{\text{v}}+\frac{1.5}{3.5}=-\frac{1-1.5}{5}$

$\Rightarrow\frac{1}{\text{v}}=\frac{1}{10}-\frac{3}{7}\Rightarrow\text{v}=\frac{-70}{23}=-3\text{cm}$ (inside the sphere).

$\Rightarrow$ Image will be formed, 2cm left to centre.

2. Image seen from right:

u = -(5 + 1.5) = -6.5cm

R = -5cm

$\therefore \ \frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$

$\Rightarrow\frac{1}{\text{v}}+\frac{1.5}{6.5}=\frac{1-1.5}{-5}$

$\Rightarrow\frac{1}{\text{v}}=\frac{1}{10}-\frac{3}{13}$

$\Rightarrow \frac{13-30}{130}$

$\Rightarrow\text{v}=\frac{-130}{17}=-7.65\text{cm}$ (inside the sphere).

$\Rightarrow$ Image will be formed, 2.65cm left to centre.

$\mu=-\infty, \ \mu_1=1, \ \mu_2=?$

1. When focused on the surface, v = 2r, R = r

So, $\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$

$\Rightarrow\frac{\mu_2}{2\text{r}}=\frac{\mu_2-1}{\text{r}}$

$\Rightarrow \frac{\mu_2}{2\text{r}}-\Big(\frac{1}{-\infty}\Big)=\frac{\mu_2-1}{\text{r}}$

$\Rightarrow\mu_2=2\mu_2-2\Rightarrow\mu_2=2$

2. When focused at centre, u = r₁, R = r

So, $\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$

$\Rightarrow\frac{\mu_2}{\text{R}}=\frac{\mu_2-1}{\text{r}}\Rightarrow\mu_2=\mu_2-1.$

This is not possible.

So, it cannot focus at the centre

$\text{m}=\frac{\text{v}}{\text{u}}=\frac{200}{3.5}\Rightarrow\text{u}=17.5\text{cm}$

$\Rightarrow\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}​​$

$\Rightarrow\frac{1}{\text{v}}-\frac{1}{-\text{u}}=\frac{1}{\text{f}}​​\Rightarrow \frac{1}{1000}+\frac{1}{17.5}=\frac{1}{\text{f}}$

$\Rightarrow\text{f}=17.19\text{cm}.$