Physics 4 Marks Questions

$\sin\text{i}_{c} = \frac{1}{\mu}_{mg} = \frac{\mu_m}{\mu_{g}}$
$\Rightarrow \mu_{m} = \mu_{g}\sin\text{i}_{c}$

$=1.5\times \frac{\sqrt{3}}{2}(\text{i}_{c} = 60^{\circ})$

$ = 1.299 \simeq1.3$

Alternate Answer

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From the diagram:

$\angle\text{i} = \angle\text{𝑁𝑂𝑀} +\angle\text{ 𝑁𝐶𝑀}$

$\angle\text{𝑟} = \angle \text{𝑁𝐶𝑀} − \angle \text{ 𝑁𝐼𝑀}$

By Snell‟s law, $\text{𝑛}_{1} \sin\text{ 𝑖} = \text{𝑛}2 \sin\text{ 𝑟}$

Substituting for i and r. and simplifying, we get

$\frac{\text{𝑛}_{1}}{\text{𝑂𝑀}}+\frac{\text{𝑛}_{2}}{\text{𝑀𝐼}}=\frac{\text{𝑛}_{2} − \text{𝑛}_{1}}{\text{MC}} $

Substituting values of OM , MI and MC

$\frac{\text{𝑛}_{2}}{\text{𝜐}}−\frac{\text{𝑛}_{1}}{\text{𝑢}}=\frac{\text{𝑛}_{2}−\text{n}_{1}}{\text{R}}$

2. Condition for real image: ν is positive

$\therefore\frac{\text{n}_{2}}{\text{v}} > 0$

From the derived relation , we have

$\frac{\text{n}_{1}}{|\text{u}|}<\frac{\text{n}_{2}-\text{n}_{1}}{\text{R}}$

$\therefore|\text{u|} > \frac{\text{n}_{1}\text{R}}{\text{n}_{2} - \text{n}_{1}}$.

For the left glass lens,

v = -40cm,

u = -25cm 

$\therefore \frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{-40}-\frac{1}{-25}=\frac{1}{25}-\frac{1}{40}=\frac{3}{200}$

$\Rightarrow \text{f}=\frac{200}{3}\text{cm}$

For the right glass lens,

v = -100cm,

u = -25cm

$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{-100}-\frac{1}{-25}=\frac{1}{100}=\frac{3}{100}$

$\Rightarrow\text{f}=\frac{100}{3}\text{cm}$

1. For an astronomical telescope, the eye piece lens should have smaller focal length. So, she should

use the right lens $\big(\text{f}=\frac{100}{3}\text{cm}\big)$ as the eye piece lens.

2. With relaxed eye, (normal adjustment)

$\text{f}_0=\frac{200}{3}\text{cm}$

$\text{f}_\text{e}=\frac{100}{3}\text{cm}$

magnification $=\text{m}=\frac{\text{f}_0}{\text{f}_\text{e}}=\frac{\big(\frac{200}{3}\big)}{\big( \frac{100}{3}\big)}=2$

1.  (c) 20cm

Explanation:

$\frac{1}{\text{f}}=(\mu-1)(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2})$

For equiconvex lens, R₁ = R, R₂ = -R

$\frac{1}{20}=(\frac{3}{2}-1)(\frac{2}{\text{R}})=\frac{1}{\text{R}}$

R = 20cm

2.  (a) Greater than air but less than water.

Explanation:

When a lens is immersed in a medium whose refractive index is greater than that of the lens, its nature changes. Here the lens changes its nature when immersed in water it means its refractive index is less than th at of water.

3.  (a) $\frac{1}{\text{f}(\text{n}-1)}$

Explanation:

According to lens maker's formula

$\frac{1}{\text{f}}=(\text{n}-1)(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2})\therefore(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2})=\frac{1}{\text{f}(\text{n}-1)}$

4.  (d) $\frac{7}{4}$

Explanation:

$\frac{1}{\text{f}_\text{m}}=(\frac{\mu_\text{g}}{\mu_\text{m}}-1)(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2})$

The given lens would behave as concave when f_m becomes negative, for which $\mu_\text{m}>\mu_\text{g}.$

5. (c) $\frac{5}{3}$

Explanation:

Here, R₁ = 20cm, R₂ = -40cm, = 20cm

Using lens maker's formula we get,

 $\frac{1}{20}=(\mu-1)(\frac{1}{20}+\frac{1}{40});\frac{1}{20}=(\mu-1)\frac{3}{40}\Rightarrow\mu=\frac{5}{3}$ 

1. (d) All of these.

Explanation:

Total internal reflection is the basis for following phenomenon:

1. Sparkling of diamond.
2. Optical fibre communication.
3. Instrument used by doctors for endoscopy.

2.  (d) Is incident at an angle greater than the critical angle.

Explanation:

Total internal reflection (TlR) is the phenomenon that involves the reflection of all the incident light off the boundary. TlR only takes place when both of the following two conditions are met: The light is in the more denser medium and approaching the less denser medium.

The angle of incidence is greater than the critical angle.

3.  (c) 90º

Explanation:

If incidence of angle, i = critical angle C, then angle of refraction, r = 90º

4. (b) n₁ > n₂

Explanation:

In optical fibres, core is surrounded by cladding, where the refractive index of the material of the core is higher than that of cladding to bound the light rays inside the core.

5. (b) 1.5 × 10⁸ms^-1

Explanation:

From Snell's law, $\sin\text{C}=\ _1\text{n}_2=\frac{\text{v}_1}{\text{v}_2}$

Where, C = critical angle = 30º and v₁ and v₂ are speed of light in medium and vacuum, respectively. We know that, v₂ = 3 × 10⁸ms^-1

$\therefore\sin30^\circ=\frac{\text{v}_1}{3\times10^8}$

$\Rightarrow\text{v}_1=3\times10^8\times\frac{1}{2}$

$\Rightarrow\text{v}_1=1.5\times10^8\text{ms}^{-1}$

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

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. (a) 41.8

Explanation:

$\sin\text{C}=\frac{1}{\mu}=\frac{1}{\frac{3}{2}}=\frac{2}{3}=0.6667$

$\text{C}=\sin^{-1}(0.6667)=41.8^{\circ}$

2.  (c) $\frac{4}{3}$

Explanation:

$\text{C}\mu=\frac{1}{\sin\text{C}}=\frac{1}{\sin48.6}=\frac{1}{0.75}=\frac{4}{3}$

3.  (c) 48º

Explanation:

From $\mu=\frac{1}{\sin\text{C}},\sin\text{C}=\frac{1}{\mu}$

As, $\mu_\text{v}>\mu_\text{r}\therefore\text{C}_\text{v}<\text{C}_\text{r}$

The correct alternative may be (c).

4.  (b) Difference between apparent and real depth of a pond.

Explanation:

Difference between apparent and real depth of a pond is due to refraction. Other three are due to total internal reflection.

5. (c) Greater than $\theta_2$

Explanation:

As, $^\text{w}{\mu}_\text{g}<\ ^\text{a}\mu_\text{w}<\ ^\text{a}\mu_\text{g};\ \therefore\theta>\theta_2>\theta_1$

1. (b) 60cm right of lens.

Explanation:

Each half lens will from an image in the same plane. The optic axes of the lenses are displaced,

$\frac{1}{\text{v}}-\frac{1}{(-30)}=\frac{1}{20};\text{v}=60\text{cm}$

2. (a) 26.7cm

Explanation:

Here f₁ = 20cm; f₂ = ?

F = 80cm

As, $\frac{1}{\text{f}_1}+\frac{1}{\text{f}_2}=\frac{1}{\text{F}}\Rightarrow\frac{1}{\text{f}_2}=\frac{1}{\text{F}}-\frac{1}{\text{f}_1}$

$\frac{1}{\text{f}_2}=\frac{1}{80}-\frac{1}{20}=\frac{-3}{80}$

$\text{f}_2=\frac{-80}{3}=-26.7\text{cm}$

3. (b) Diverging lens.

Explanation:

The bubble behaves libe a diverging lens.

4. (b) Convex lens.

Explanation:

Convex lens is used in magnifying glass.

5. (d) Between F and optical centre.

1. (a) 5cm, 35cm

Explanation:

$\text{m}=\frac{\text{f}_0}{\text{f}_\text{e}}=7$

$\text{f}_0=7\text{f}_e$

In normal adjustment, distance between the lenses,

$\text{f}_0+\text{f}_\text{e}=40$

$7\text{f}_0+\text{f}_\text{e}=40\Rightarrow\text{f}_\text{e}=\frac{40}{8}=5\text{cm}$

$\text{f}_0=7\text{f}_\text{e}=7\times5=35\text{cm}$

2.  (d) 20cm

Explanation:

$\text{m}=-10;\text{L}=22\text{cm}$

As, $\text{m}=\frac{-\text{f}_0}{\text{f}_e}\Rightarrow-10=-\frac{\text{f}_0}{\text{f}_\text{e}}$

$\text{f}_0=10\text{f}_\text{e}$

As, $\text{L}=\text{f}_0+\text{f}_\text{e}$

$22=10\text{f}_\text{e}+\text{f}_e=11\text{f}_\text{e}$

or $\text{f}_\text{e}=\frac{22}{11}=2\text{cm}$

$\text{f}_0=10\text{f}_\text{e}=20\text{cm}$

3.  (d) Large focal length.

Explanation:

Objective lens has larger focal length than eye-piece.

4.  (d) Astronomical telescope.

Explanation:

Astronomial telescope is used to see stars, sun etc.

5. (c) f₀ << f_e

1. (a) Red

Explanation:

Angle of deviation is minimum for the red colour.

2.  (a) All colours have same speed.

Explanation:

In vacuum all colours have same speed, because there is no dispersion of light in vacuum.

3.  (c) 90º

Explanation:

The deviation is maximum when angle is 90º.

4. (a) 3.864º

Explanation:

$\text{A}=6^\circ;\mu=1.644$

$\text{f}(\mu-1)\text{A}$

$\text{f}=(1.644-1)6=0.644\times6$

$\delta=3.864^\circ$

5. (d) 40º

Explanation:

$\text{i}_1=50^\circ;\text{A}=60^\circ,\delta^\text{m}=?$

$\text{A}+\delta_\text{m}=\text{i}_1+\text{i}_2=50^\circ+50^\circ=100^\circ$

$\delta_\text{m}=100^\circ=\text{A}=100-60^\circ=40^\circ$

1.  (d) 30cm

Explanation:

Here, $\mu=1.5;\text{R}_1=30\text{cm},\ \text{R}_2=-30\text{cm}$

As, $\frac{1}{\text{f}}=(\mu-1)[\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}]$

$=(1.5-1)[\frac{1}{30}-\frac{1}{-30}]=-0.5\times\frac{2}{30}=\frac{-1}{30}$

$\text{f}=-30\text{cm}$

2.  (a) 1.5

Explanation:

Here, $\text{f}=12;\text{R}_1=10\text{cm},\ \text{R}_2=-15\text{cm}$

As,$\frac{1}{\text{f}}=(\mu-1)[\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}]$

$\frac{1}{12}=(\mu-1)[\frac{1}{10}+\frac{1}{15}]$

$\mu=1.5$

3.  (d) Change in the focal length of eye-lens.

Explanation:

The eye-lens is surrounded by a different medium than air. This will change the focal length of the eye-lens. The eye cannot accommodate all images as it would do in air.

4. (b) 40cm

Explanation:

$\frac{1}{\text{f}}=(1.5-1)(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2})$

and $\frac{1}{\text{f}_\text{w}}=(\frac{1.5}{1.33}-1)(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2})$

$\frac{\text{f}_\text{w}}{\text{f}}=\frac{0.5\times1.33}{0.17}=4$

$\text{f}_\text{w}=4\text{f}=4\times10=40\text{cm}$

5. (d) Have refractive index exactly matching with that of the surrounding fluid.

Explanation:

If the refractive index of two media are same, the surface of separation does not produce refraction or reflection which helps in visibility. 

1. (d) All of these.

Explanation:

Refractive indexofa medium depends upon nature and temperature of the medium, wavelength of light.

2.  (a) 1.33

Explanation:

Here, $\text{v}=5\times10^{14}\text{Hz};\lambda=450\times10^{-9}\text{m}$

$\text{c}=3\times10^8\text{ms}^{-1}$

Refractive index of the liquid,

$\mu=\frac{\text{c}}{\text{v}}=\frac{\text{c}}{\text{v}\lambda}=\frac{3\times10^8}{5\times10^{14}\times450\times10^{-9}}$

$\mu=1.33$

3.  (b) $\sin^-1 (0.58)$

Explanation:

Here, $\text{i}=60^\circ;\mu=1.5$

By snell's law, $\mu=\frac{\sin\text{i}}{\sin\text{r}}$

$\sin\text{r}=\frac{\sin\text{i}}{\mu}=\frac{\sin60^\circ}{1.5}=\frac{0.866}{1.5}$

$\sin\text{r}=0.5773$ or $\text{r}=\sin^{-1}(0.58)$

4. (c) 6cm

Explanation:

As object is at the centre of the sphere, the image must be at the centre only.

$\therefore$ Distance of virtual image from centre of sphere = 6cm.

5. (c) The speed is different.

Explanation:

Speed of light in second medium is different than that in first medium.

1. (d) Increases.
2. (a) 2.5D

Explanation:

P = P₁ + P₂ = 1.5 + 1.0 = 2.5D

3. (b) 20cm

Explanation:

$\text{p}=\frac{1}{\text{p}}=\frac{1}{5}\text{m}=+20\text{cm}$

4. (a) -10D

Explanation:

$\text{p}=\text{p}_1+\text{p}_2=\frac{1}{\text{f}_1}+\frac{1}{\text{f}_2}$

$\frac{100}{10}+\frac{100}{-5}=-10\text{D}$

5. (b) Diverging in nature.

Explanation:

$\text{p}=\text{p}_1+\text{p}_2=\frac{100}{\text{f}_1}+\frac{100}{\text{f}_2}$

$\frac{100}{25}+\frac{100}{-25}=-1\text{D}$

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.

1. (b) 5cm

Explanation:

Here, f₀ = 2.0, f_e = 6.25cm, u₀ = ?

When the final image is obtained at the least distance of distinct vision:

v_e = -25cm

As $\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{}_\text{e}}=\frac{1}{\text{f}_\text{e}}$

$\therefore\frac{1}{\text{u}_\text{e}}\frac{1}{\text{v}_\text{e}}\frac{1}{\text{f}_\text{e}}=\frac{1}{-25}-\frac{1}{6.25}$

$=\frac{-1-4}{25}=\frac{-5}{25}=-\frac{1}{5}$

or $\text{u}_\text{e}=-5\text{cm}$

2.  (b) 2.5cm

Explanation:

Distance between objective and eye-piece = 15cm

$\therefore$ Distance of the image from objective is $\text{v}_0=15-5=10\text{cm}$

$\therefore\frac{1}{\text{u}_0}=\frac{1}{\text{v}_0}-\frac{1}{\text{f}_0}=\frac{1}{\text{10}}-\frac{1}{\text{2}}=\frac{1-5}{\text{10}}=-\frac{2}{5}$

or $\text{u}_0=-\frac{5}{2}=-2.5\text{cm}$

$\therefore$ Distance of object from objective = 2.5cm

3.  (a) 20

Explanation:

Magnifying power,

$\text{m}=\text{m}_0\times\text{m}_\text{e}=\frac{\text{v}_0}{\text{u}_0}(1+\frac{\text{D}}{\text{fe}})=\frac{10}{2.5}(1+\frac{25}{6.25})=20$

4. (a) Real, inverted and magnified.

Explanation:

The intermediate image formed by the objective of a compound microscope is real, inverted and magnified.

5. (d) Focal lengths of both objects and eye-piece are decreased.