2 Marks Questions

Question 512 Marks

The eyepiece of an astronomical telescope has a focal length of 10cm. The telescope is focused for normal vision of distant objects when the tube length is 1.0m. Find the focal length of the objective and the magnifying power of the telescope.

Answer

For the given astronomical telescope in normal adjustment,

F_e = 10cm, L = 1m = 100cm

S0, f₀ = L - f_e = 100 - 10 = 90cm

and, magnifying power $=\frac{\text{f}_0}{\text{f}_\text{e}}=\frac{90}{10}=9$

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Question 522 Marks

"Monochromatic light should be used to produce pure spectrum". Comment on this statement.

Answer

No, monochromatic light cannot be used to produce a pure spectrum. A spectrum is produced when a light of different wavelengths is deviated through different angles and gets separated. Monochromatic light, on the other hand, has a single wavelength.

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Question 532 Marks

A compound microscope forms an inverted image of an object. In which of the following cases it is kely to create difficulties?

Looking at small germs.
Looking at circular spots.
Looking at a vertical tube containing some water.

Answer

Looking at a vertical tube containing some water

Explantion:

If the experimentalist is looking at a vertical tube containing some water, he has to be careful, as the lower meniscus will appear as upper.

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Question 542 Marks

A nearsighted person cannot clearly see beyond 200cm. Find the power of the lens needed to see objects at large distances.

Answer

For the near sighted person,

$\text{u}= \infty$ and $\text{v}=-200\text{cm}=-2\text{m}$

So, $\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{-2}-\frac{1}{\infty}$

$\frac{1}{\text{f}}=-\frac{1}{2}=-0.5$

So, power of the lens is -0.5D

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Question 552 Marks

A farsighted person cannot see objects placed closer to 50cm. Find the power of the lens needed to see the objects at 20cm.

Answer

For the far sighted person,

u = -20cm, v = -50cm

from lens formula $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$

$\frac{1}{\text{f}}=\frac{1}{-50}-\frac{1}{-20}=\frac{1}{-20}-\frac{1}{50}=\frac{3}{100}$

$\Rightarrow\text{f}=\frac{100}{3}\text{cm}=\frac{1}{3}\text{m}$

So, power of the lens $=\frac{1}{\text{f}}=3\text{ Diopter}$

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Question 562 Marks

A light ray is incident normally on the face AB of a right-angled prism ABC $(\mu=1.50)$ as shown in figure. What is the largest angle $\phi$ for which the light ray is totally reflected at the surface AC?

Answer

Let $\theta_\text{c}$ be the critical angle for the glass
$\frac{\sin\theta_{\text{c}}}{\sin90^{\circ}}=\frac{1}{\text{x}}\Rightarrow\sin\theta_{\text{c}}=\frac{1}{1.5}=\frac{2}{3}\Rightarrow\theta_{\text{c}}=\sin^{-1}\Big(\frac{2}{3}\Big)$
From figure, for total internal reflection, $90^{\circ}-\phi>\theta_{\text{c}}$
$\Rightarrow\phi<90^{\circ}-\theta_{\text{c}}\Rightarrow\phi<\cos^{-1}\Big(\frac{2}{3}\Big)$
So, the largest angle for which light is totally reflected at the surface is $\cos^{-1}\Big(\frac{2}{3}\Big).$

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Question 572 Marks

A 1cm object is placed perpendicular to the principal axis of a convex mirror of focal length 7.5cm. Find its distance from the mirror if the image formed is 0.6cm in size.

Answer

$\text{m}=-\frac{\text{v}}{\text{u}}=0.6$ and $\text{f}=7.5\text{cm}=\frac{15}{2}\text{cm}$
From mirror equation,
$\Rightarrow\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{0.6\text{u}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow \frac{5}{3\text{u}}-\frac{1}{\text{u}}=\frac{5}{15}$
$\Rightarrow\text{u}=5\text{cm}$

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Question 582 Marks

An optical fibre $(\mu=1.72)$ is surrounded by a glass coating $(\mu=1.50).$ Find the critical angle for total internal/ reflection at the fibre-glass interface.

Answer

For calculation of critical angle,
$\frac{\sin\text{i}}{\sin\text{r}}=\frac{\mu_2}{\mu_1}$
$\Rightarrow\frac{\sin\theta_\text{C}}{\sin90^\circ}=\frac{15}{1.72}$
$=\frac{75}{86}$
$\Rightarrow\theta_\text{C}=\sin^{-1}\Big(\frac{75}{86}\Big)$

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Question 592 Marks

A point source is placed at a depth h below the surface of water (refractive index = μ).

Show that light escapes through a circular area on the water surface with its centre directly above the point source.
Find the angle subtended by a radius of the area on the source.

Answer

Let, x = radius of the circular area

$\frac{\text{x}}{\text{h}}=\tan\theta_\text{C}$ (where C is the critical angle)

$\Rightarrow\frac{\text{x}}{\text{h}}=\frac{\sin\theta_\text{C}}{\sqrt{1-\sin^2\theta_\text{C}}}=\frac{\frac{1}{\mu}}{\sqrt{1-\frac{1}{\mu^2}}} \ \Big(\because \ \sin\theta_\text{C}=\frac{1}{\mu}\Big)$

$\Rightarrow\frac{\text{x}}{\text{h}}=\frac{1}{\sqrt{\mu^2-1}}$ or $\text{x}=\frac{\text{h}}{\sqrt{\mu^2-1}}$

So, light escapes through a circular area on the water surface directly above the point source.

Angle subtained by a radius of the area on the source, $\theta_\text{C}=\sin^{-1}\Big(\frac{1}{\mu}\Big)$

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Question 602 Marks

Light falls from glass $(\mu=1.5)$ to air. Find the angle of incidence for which the angle of deviation is 90°.

Answer

Since, $\mu=1.5,$ Critial angle $=\sin^{-1}\Big(\frac{1}{\mu}\Big)=\sin^{-1}\Big(\frac{1}{1.5}\Big)=41.8^{\circ}$
We know, the maximum attainable deviation in refraction is (90° - 41.8°) = 47.2°
So, in this case, total internal reflection must have taken place.
In reflection,
Deviation = 180° - 2i = 90°
⇒ 2i = 90° ⇒ i = 45°.

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Question 612 Marks

A concave mirror and a converging lens have the same focal length in air. Which one of the two will have greater focal length when both are immersed in water?

Answer

Converging lens; the focal length of a spherical mirror remains unaffected.
For converging lens $\frac{1}{\text{f}}=\Big(\frac{\text{n}_1}{\text{n}_2}-1\Big)\Big(\frac{1}{\text{R}_2}-\frac{1}{\text{R}_2}\Big)$
When it is immersed in water n₂ (in water) > n₂ (air)
$\Big(\frac{\text{n}_1}{\text{n}_2}-1\Big)$ decreases hence focal length of converging lens increases in water.

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Question 622 Marks

A single lens is mounted in a tube. A parallel beam enters the tube and emerges out of the tube as a divergent beam. Can you say with certainty that there is a diverging lens in the tube?

Answer

No,

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Question 632 Marks

Can you ever have a situation in which a light ray goes undeviated through a prism?

Answer

Yes.

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Question 642 Marks

A spherical surface of radius 30cm separates two transparent media A and B with refractive indices 1.33 and 1.48 respectively. The medium A is on the convex side of the surface. Where should a point object be placed in medium A so that the paraxial rays become parallel after refraction at the surface?

Answer

Since, paraxial rays become parallel after refraction i.e. image is formed at $\infty.$
$\text{v}=\infty, \ \mu_1=1.33, \ \text{u}=?, \ \mu_2=1.48, \ \text{R}=30\text{cm}$
$\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$
$\Rightarrow\frac{1.48}{\infty}-\frac{1.33}{\text{u}}=\frac{1.48-1.33}{30}\Rightarrow-\frac{1.33}{\text{u}}-\frac{0.15}{30}$
$\Rightarrow\text{u}=-266.0\text{cm}$
$\therefore$ Object should be placed at a distance of 266cm from surface (convex) on side A.

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Question 652 Marks

Redraw the diagram given below and mark the position of the centre of curvature of the spherical mirror used in the given set up.

Answer

If the object is in between focus 'F' and centre of curvature 'C', image would be beyond the centre of curvature, inverted real and magnified.

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Question 662 Marks

Can the dispersive power $\omega=\frac{\mu_\text{v}-\mu\text{r}}{\mu-1}$ be negative? What is the sign of co if a hollow prism is immersed into water?

Answer

No, it cannot be negative, as the refractive index for violet light is always greater than that for red light. Also, refractive index is inversely proportional to $\lambda^2$. The sign of wwill be positive, as $\mu$ is still greater than 1 and es $\mu_\text{v}>\mu_\text{r}.$

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Question 672 Marks

The minimum deviations suffered by red, yellow and violet beams passing through an equilateral transparent prism are 38.4°, 38.7° and 39.2° respectively. Calculate the dispersive power of the medium.

Answer

Given that, $\delta_\text{r}=38.4^\circ,\delta_\text{y}=38.7^\circ\ \text{and}\ \delta_\text{v}=39.2^\circ$
Dispersive power $=\frac{\mu_\text{v}-\mu_\text{r}}{\mu_\text{y}-1}=\frac{(\mu_\text{v}-1)-(\mu_\text{r}-1)}{(\mu_\text{y}-1)}$ $=\frac{\Big(\frac{\delta_\text{v}}{\text{A}}\Big)-\Big(\frac{\delta_\text{r}}{\text{A}}\Big)}{\Big(\frac{\delta_\text{v}}{\text{A}}\Big)}\ [\because\delta=(\mu-1)\text{A}]$
$=\frac{\delta_\text{v}-\delta_\text{r}}{\delta_\text{y}}=\frac{39.2-3.84^\circ}{38.7}=0.0204$

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Question 682 Marks

The refractive index of a material of a concave lens is n₁. It is immersed in a medium of refractive index n₂ A parallel beam of light is incident on the lens. Trace the path of emergent rays when (i) n₂= n₁ (ii) n₂ > n₁ (iii) n₂ < n₁.
$\frac{1}{\text{f}}=\Big(\frac{\text{n}_1}{\text{n}_2}-1\Big)\Big(\frac{1}{\text{R}_2}-\frac{1}{\text{R}_2}\Big)$

Answer

$\text{For} \text{ n}_1 = \text{n}_2\text{ f} = \infty$
$\text{For} \text{ n}_1 < \text{n}_2 \text{ f} > 0$
$\text{For} \text{ n}_1 > \text{n}_2 \text{ f} < 0$
The path of rays in three cases is shown in fig.

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Question 692 Marks

Find the maximum angle of refraction when a light ray is refracted from glass $(\mu=1.50)$ to air.

Answer

From the definition of critical angle, if refracted angle is more than 90°, then reflection occurs, which is known as total internal reflection.
So, maximum angle of refraction is 90°.

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Question 702 Marks

If an object far away from a convex mirror moves towards the mirror, the image also moves. Does it move faster, slower or at the same speed as compared to the object?

Answer

Slower $\text{V}_{\text{img}}=\text{m}^2\text{v}_0$ [m < 1 as far away]

It objects moves from infinity to 2f as distance moved by object is more in same time hence velocity of object is more.

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Question 712 Marks

Why is there no dispersion in the light refracted through a rectangular glass slab?

Answer

It can be considered to be equivalent to two prisms in reverse way. In the case of a rectangular glass slab the rays of all colours combine together in the reverse prism. Hence, there is no dispersion.

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Question 722 Marks

An object is placed at the principal focus of a concave lens of focal length f. Where will its image be formed?
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\Rightarrow\frac{1}{\text{v}}=\frac{1}{\text{f}}+\frac{1}{\text{u}}$

Answer

Here u = -f and for a concave lens f = -f
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\Rightarrow\frac{1}{\text{v}}=\frac{1}{\text{f}}+\frac{1}{\text{u}}$
That image will be formed between optical centre and focus of lens; towards the side of the object.

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Question 732 Marks

The focal length of a convex lens made of glass is 20cm. What will be its new focal length when placed in a medium of refractive index 1.25?

Answer

$\frac{1}{\text{f}}=(\mu-1)\Big[\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big]$
$\Rightarrow\frac{1}{20}=\frac{1}{4}\Big[\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big]$
$\frac{1}{\text{f}}=\frac{1}{4}\Big[\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big]$
$\Rightarrow\text{f}'=40\text{cm}$

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Question 742 Marks

The equation of refraction at a spherical surface is $\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}-=\frac{\mu_2-\mu_1}{\text{R}}.$ Taking $\text{R}=\infty,$ show that this equation leads to the equation $\frac{\text{Real depth}}{\text{Apparent depth}}=\frac{\mu_2}{\mu_1}$ for refraction at a plane surface.

Answer

$\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}-=\frac{\mu_2-\mu_1}{\infty}$
$\frac{\mu_2}{\text{v}}=\frac{\mu_1}{\text{u}}$
$\Rightarrow\frac{\mu_2}{\mu_1}=\frac{\text{v}}{\text{u}}.$

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Question 752 Marks

Light is incident from glass $(\mu=1.5)$ to air. Sketch the variation of the angle of deviation $\delta$ with the angle of incident i for 0 < i < 90°.

Answer

Refractive index of glass $\mu_{\text{g}}=1.5$
Given, 0° < i < 90°
Let, $\theta_\text{C}$ → Critical angle.
$\frac{\sin\theta_\text{C}}{\sin\text{r}}=\frac{\mu_{\text{a}}}{\mu_{\text{g}}}$
$\Rightarrow\frac{\sin\theta_\text{C}}{\sin90^{\circ}}=\frac{1}{1.5}=0.66$
$\Rightarrow\text{C}=40^{\circ}48''$
The angle of deviation due to refraction from glass to air increases as the angle of incidence increases from 0° to 40°48''. The angle of deviation due to total internal reflection further increases for 40°48'' to 45° and then it decreases.

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Question 762 Marks

A right-angled crown glass prism with critical angle 41° is placed before an object, PQ in two positions as shown in the figures (i) and (ii). Trace the paths of the rays from P and Q passing through the prisms in the two cases.

Answer

The formation of images is shown in figures (i) and (ii).

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Question 772 Marks

If three identical prisms are combined, is it possible to pass a beam that emerges undeviated? Undispersed?

Answer

No, it is not possible even when prisms are be combined with their refractive angle reversed with respect to each other. There will be at least a net deviation and dispersion equal to the dispersion and deviation produced by a single prism.

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Question 782 Marks

Can a plane mirror ever form a real image?

Answer

It object is virtual image is real.

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Question 792 Marks

A concave lens is placed in water. Will there be any change in focal length? Give reason.

Answer

Focal length of lens in water $\text{f}_{\text{w}}=\frac{\text{n}_{\text{g}}-1}{\frac{\text{n}_{\text{g}}-1}{\text{nw}}-1}$
$\text{As n}_{\text{g}}>\text{n}_{\text{w}},\frac{\text{n}_{\text{g}}}{\text{n}_{\text{w}}}>1,\text{ so }\text{f}_{\text{w}}>\text{f}_{\text{a}}$

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Question 802 Marks

Suggest a method to produce a rainbow in your house.

Answer

A rainbow can be produced using a prism. Another way of producing a rainbow is to dip a mirror inside water, keeping it inclined along the wall of a tumbler. The light coming from water after reflecting from the mirror will give a rainbow.

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Question 812 Marks

How does the power of a convex lens vary, if the incident red light is replaced by violet light?

Answer

Power of a lens increases if red light is replaced by violet light because.
$\text{P}=\frac{1}{\text{f}}=(_{\text{a}\text{n}_{\text{g}}-1})\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_1}\Big)$ and refractive index is maximum for violet light in visible region of spectrum.

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Question 822 Marks

A telescope has been adjusted for relaxed eye. You are asked to adjust it for least distance of distinct vision, then how will you change the distance between two lenses?

Answer

For relaxed eye, $\text{L} = \text{f}_0 +\text{f}_{\text{e}}$
For least distance of distinct vision $\text{L}' = \text{f}_0 + \text{u}_{\text{e}}, \text{u}_{\text{e}} < \text{f}_{\text{e}}$
Therefore, L' < L, that is, the distance will be decreased.

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Question 832 Marks

A simple microscope using a single lens often shows coloured image of a white source. Why?

Answer

A simple microscope consists of a single convex lens. Sometimes due to chromatic and spherical aberrations, the image of a white source seems coloured at the corners of the lens and somewhere in between.

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Question 842 Marks

Can a virtual image be photographed by a camera?

Answer

Yes, when you stand in front of plane mirror image is virtual and can be photographed.

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Question 852 Marks

A convex lens made of a material of refractive index n₁ is kept in a medium of refractive index n₂. Parallel rays of light are incident on the lens. Complete the path of rays of light emerging from the convex lens if: (i) n₁ > n₂ (ii) n₁ = n₂ (iii) n₁< n₂.
$\frac{1}{\text{f}}=\Big(\frac{\text{n}_1}{\text{n}_2}-1\Big)\Big(\frac{1}{\text{R}_2}+\frac{1}{\text{R}_2}\Big)$

Answer

In case (i) n₁ > n₂, the lens behaves as convergent lens.
In case (ii) n₁= n₂, the lens behaves as a plane plate.
In case (iii) n₁< n₂, the lens behaves as a divergent lens.

The path of rays in all the three cases is shown in fig.

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Question 862 Marks

A simple microscope is rated 5 for a normal relaxed eye. What will be its magnifying power for a relaxed farsighted eye whose near point is 40cm?

Answer

The simple microscope has magnification of 5 for normal relaxed eye (D = 25cm).

Because, the eye is relaxed the image is formed at infinity (normal adjustment)

So, $\text{m}=5=\frac{\text{D}}{\text{f}}=\frac{25}{\text{f}}\Rightarrow\text{f}=5\text{cm}$

For the relaxed farsighted eye, $\text{D} = 40\text{cm}$

So, $\text{m}=\frac{\text{D}}{\text{f}}=\frac{40}{5}=8$

So, its magnifying power is 8X.

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Question 872 Marks

A convex lens (n = 1.5) of focal length fa is immersed.

In water n = 1.33 and.
In carbon disulphide n = 1.6, how does the lens behave in the two cases?

Answer

When lens is immersed in water, it behaves as a convex lens but its focal length will increase.
When convex lens is immersed in carbon-disulphide, it will behave as a concave lens.

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Question 882 Marks

The equation $\omega=\frac{\mu_\text{v}-\mu_\text{r}}{\mu-1}$ asderived for a prism having small refracting angle. Is it also valid for a prism of large refracting angle? Is it also valid for a glass slab or a glass sphere?

Answer

Dispersive power depends on angular deviation, and angular deviation is valid only for a small refracting angle and a small angle of incidence. Therefore, dispersive power is not valid for a prism of large refracting angle. It is also not valid for a glass slab or a glass sphere, as it has a large refracting angle.

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Question 892 Marks

An extended object is placed at a distance of 5.0cm from a convex lens of focal length 8.0cm.

Draw the ray diagram (to the scale) to locate the image and from this, measure the distance of the image from the lens.
Find the position of the image from the lens formula and see how close the drawing is to the correct result.

Answer

Given, u = -5cm, f = 8cm

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

$\Rightarrow\frac{1}{8}-\frac{1}{5}=\frac{-3}{40}$

$\Rightarrow\text{v}=-13.3\text{cm}$ (virtual image).

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Question 902 Marks

A Galilean telescope is 27cm long when focused to form an image at infinity. If the objective has a focal length of 30cm, what is the focal length of the eyepiece?

Answer

For the given Galilean telescope, (When the image is formed at infinity)

f₀ = 30cm, L = 27cm

Since L = f₀ - |f_e|

[Since, concave eyepiece lens is used in Galilean Telescope]

⇒ f_e = f₀ - L = 30 - 27 = 3cm

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Question 912 Marks

The refractive index of a material changes by 0.014 as the colour of the light changes from red to violet. A rectangular slab of height 2.00cm made of this material is placed on a newspaper. When viewed normally in yellow light, the letters appear 1.32cm below the top surface of the slab. Calculate the dispersive power of the material.

Answer

Given that, $\mu_\text{v}-\mu_\text{r}=0.014$

$\text{Again,}\ \mu_\text{y}=\frac{\text{Real depth}}{\text{Apparent depth}}=\frac{2.00}{1.30}=1.515$

So, dispersive power $=\frac{\mu_\text{v}-\mu_\text{r}}{\mu_\text{y}-1}=\frac{0.014}{1.515-1}=0.027$

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Question 922 Marks

A person wears glasses of power -2.5D. Is the person farsighted or nearsighted? What is the far point of the person without the glasses?

Answer

The person wears glasses of power -2.5D

So, the person must be near sighted

$\text{u}=\infty,$ $\text{v}=\text{far point,}$ $\text{f}=\frac{1}{-2.5}=-0.4\text{m}=-40\text{cm}$

Now, $\frac{1}{\text{v}}-\frac{1}{\text{v}}=\frac{1}{\text{f}}$

$\Rightarrow\frac{1}{\text{v}}=\frac{1}{\text{u}}+\frac{1}{\text{f}}=0+\frac{1}{-40}$

$\Rightarrow \text{v}=-40\text{cm}$

So, the far point of the person is 40cm.

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Question 932 Marks

The angular magnification of a system is less than one. Does it mean that the image formed is inverted?

Answer

No, angular magnification is the ratio of the angle subtended by the final image on the eye to the angle subtended by the object on the unaided eye. Its value less than one signifies reduction in the size of the image. It does not mean that the image is inverted.

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Physics 2 Marks Questions