1 Marks Question - Physics STD 12 Science Questions - Vidyadip

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22 questions · 4 auto-graded MCQ + 18 self-marked written.

Question 11 Mark

Two light sources of intensities $8cd$ and $12cd$ are placed on the same side of a photometer screen at a distance of $40\ cm$ from it. Where should a $80cd$ source be placed to balance the illuminance?

Answer

Total intensity of the $8cd$ and the $12cd$ light source is $20cd$.
Let $d$ be the distance of the $80cd$ source.
$\therefore$ Illuminance due to the $20cd$ source.
$\text{E}_1=\frac{20}{(0.4)^2}\ \dots(1)$
$\therefore$ Illuminance due to the $80cd$ source $(E_2)$ is:
$\text{E}_2=\frac{80}{\text{d}^2}\ \dots(2)$
As$, E_{1 }= E_{2}$
$\therefore\frac{20}{(0.4)^2}=\frac{80}{\text{d}^2}$
$\Rightarrow\text{d}=0.8\text{m}=80\text{cm.}$

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Question 21 Mark

A room is illuminated by an extended source. The illuminance at a particular portion of a wall can be increased by:

Moving the source.
Rotating the source.
Bringing some mirrors in proper positions.
Changing the colour of the source.

Answer

Moving the source.
Rotating the source.
Bringing some mirrors in proper positions.
Changing the colour of the source.

Explanation:

Moving the source to the middle will illuminate the surface properly because illuminance depends upon the distance from the source.
Rotating the source will also have an effect because illuminance depends upon the angle made by the normal on the surface.
Bringing mirrors to the proper position will increase illuminance at that particular portion of the wall by gathering light and focussing them at one point.
Our eyes sense some colours as bright and some colours as dull, selecting the colours near yellow will make the wall appear brighter.

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Question 31 Mark

A light source emits monochromatic light of wavelength 555nm. The source consumes 100W of electric power and emits 35W of radiant flux. Calculate the overall luminous efficiency.

Answer

Overall luminous efficiency $=\frac{\text{Total luminous flux}}{\text{Power input}}=\frac{35\times685}{100}=239.75\text{ lumen}/\text{W}$

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Question 41 Mark

Light from a point source falls on a small area placed perpendicular to the incident light. If the area is rotated about the incident light by an angle of 60°, by what fraction will the illuminance change?

Answer

The illuminance will not change.

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Question 51 Mark

A battery-operated torch is adjusted to send an almost parallel beam of light. It produces an illuminance of 40 lux when the light falls on a wall 2m away. The illuminance produced when it falls on a wall 4m away is close to:

40 lux
20 lux
10 lux
5 lux

Answer

40 lux

Explanation:
Since the beam is parallel, it will have no angular spread. So the illuminance will remain same throughout. Therefore, in this case, it will be 40 lux.

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Question 61 Mark

Mark the correct options.

Luminous flux and radiant flux have same dimensions.
Luminous flux and luminous intensity have same dimensions.
Radiant flux and power have same dimensions.
Relative luminosity is a dimensionless quantity.

Answer

Luminous flux and luminous intensity have same dimensions.
Radiant flux and power have same dimensions.
Relative luminosity is a dimensionless quantity.

Explanation:

No, luminous flux has the dimension of luminous intensity (cd/sr). The dimension of radiant flux is watt.
Yes, both have the dimension of luminous intensity, i.e. cd/sr.
Yes, both have the dimensions of power.
Yes, it is a ratio of same kind of quantities. So, it is dimensionless.

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Question 71 Mark

An electric bulb is hanging over a table at a height of 1m above it. The illuminance on the table directly below the bulb is 40 lux. The illuminance at a point on the table 1m away from the first point will be about:

10 lux
14 lux
20 lux
28 lux

Answer

14 lux

Here,
$\text{r}=\sqrt{2}$
$\tan\theta=\frac{\text{BC}}{\text{AB}}=1$
$\text{I}_\text{o}=40 \text{ lux}$
$\theta=\tan^{-1}(1)=45^0$
The illluminance is given by,
$\text{E}=\frac{\text{l}_\text{0}\cos(45^0)}{\text{r}^2}$
$=\frac{40\times{\cos(45^0})}{(\sqrt{2})^2}$
$=14 \text{ lux}$

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MCQ 81 Mark

Three light sources $A, B$ and $C$ emit equal amount of radiant energy per unit time. The wavelengths emitted by the three sources are $450\ nm, 555\ nm$ and $700\ nm$ respectively. The brightness sensed by an eye for thesources are $X_{A,} X_B$ and $X_C$ respectively. Then,

A
$X_A > X_B, X_C > X_B$
✓
$X_A > X_B, X_B > X_C$
C
$X_B > X_A, X_B > X_C$
D
$X_B > X_A, X_C > X_B$

Answer

Correct option: B.

$X_A > X_B, X_B > X_C$

Wavelength of light $B$ is $555\ nm.$ It has the highest luminosity$;$
hence$, X_B$ will be highest.
Again, $450\ nm$ is nearer to $555\ nm$ than $700\ nm.$
$\therefore 555 - 450 = 105$
But $700 - 555 = 145$
So$, X_A\ 's$ brightness will be greater than that of $X_C.$

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Question 91 Mark

A source emits 45 joules of energy in 15s. What is the radiant flux of the source?

Answer

Radiant Flux $=\frac{\text{Total energy emitted}}{\text{Time}}=\frac{45}{15\text{s}}=3\text{W}$

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Question 101 Mark

What is the luminous flux of a source emitting radio waves?

Answer

The luminous flux of a source emitting radio waves will be zero, as the luminosity of radio waves is zero.

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Question 111 Mark

The intensity produced by a long cylindrical light source at a small distance r from the source is proportional to:

$\frac{1}{\text{r}^2}$
$\frac{1}{\text{r}^3}$
$\frac{1}{\text{r}}$
None of these.

Answer

$\frac{1}{\text{r}}$

Explanation:

Let us consider two coaxial cylindrical surfaces at distances r and r' from the axis.
Let areas dA and dA' subtend the solid angle $\text{d}\omega$ at the central axis.
The height of the area element will be same, i.e. equal to dy.
Let the breath of dA be dx and that of dA' be dx'.
Now from the arcs,
${\text{dx}}=\text{rd}\theta$
$\text{dx} '=\text{r}'\text{d}\theta$
Now,
$\text{dA}'=\text{dxdy}=\text{rd}\theta\text{dy}$
$\text{dA}'=\text{dx}'\text{dy}=\text{r}'\text{d}\theta\text{dy}$
$\frac{\text{dA}}{\text{dA}'}=\frac{\text{r}}{\text{r}'}$
$\Rightarrow\frac{\text{dA}}{\text{r}}=\frac{\text{dA}^,}{\text{r}^,}=\text{d}\omega$
The luminous flux going through the solid angle dω will be:
$\text{dF} = \text{I}\text{d}\omega$
Now,
$\text{dF}=\text{I}\frac{\text{dA}}{\text{r}}$
If the surfaces are inclined at an angle $\alpha,$
$\text{dF}=\text{I}\frac{\text{dA}\cos{\alpha}}{\text{r}}$
Now, illuminance is defined as
$\text{E}=\frac{{\text{dF}}}{\text{dA}}=\text{I}\frac{\text{dA}\cos\alpha}{\text{r}}$
$\Rightarrow \text{E}\propto\frac{1}{\text{r}}$

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Question 121 Mark

An electric lamp and a candle produce equal illuminance at a photometer screen when they are placed at $80\ cm$ and $20\ cm$ from the screen respectively. The lamp is now covered with a thin paper which transmits $49\%$ of the luminous flux. By what distance should the lamp be moved to balance the intensities at the screen again?

Answer

Let $I_1$ be the intensity when placed at a distance $80\ cm$ and $I_2$ be the intensity when placed at a distance $20\ cm$ apart from the screen.
Now,$\frac{\text{I}_1}{\text{I}}=\Big(\frac{80}{20}\Big)^2=16$
According to the question, let the new distance between the lamp and the screen be $x$ such that even after covering the lamp with a thin paper the intensities at the screen is balanced.
So,$\frac{0.49\text{I}_1}{\text{I}_2}=\Big(\frac{\text{x}}{20}\Big)^2$
$\Rightarrow0.49\times16\times400=\text{x}^2$
$\Rightarrow\text{x}=56\ \text{cm}$
Thus, the lamp has to be moved by $80\ cm - 56\ cm = 24\ cm.$

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MCQ 131 Mark

A photographic plate placed at a distance of $5\ cm$ from a weak point source is exposed for $3s$. If the plate is kept at a distance of $10\ cm$ from the source, the time needed for the same exposure is:

A
$3s$
✓
$12s$
C
$24s$
D
$48s$

Answer

Correct option: B.

$12s$

Here,
$\text{d}_1=5\ cm= 0.05\text{m}$
$\text{d}_2=10\ cm=0.1\text{m}$
$\text{t}_1=3\text{s}$
$\text{t}=?$
Let the actual incident illuminance be $E_o$
Let the iluminance at $3\ cm$ distance be $E_{d1}$
Let the iluminance at $10\ cm$ distance be $E_{d2}$
$\cos\theta=1$
$\text{ E}_\text{d1}=\frac{\text{E}_\text{o}}{\text{d}_1^2}$
Now,
$\text{t}_1\alpha\frac{1}{\text{E}_\text{d1}}$
$\Rightarrow \text{t}_1=\frac{\text{k5}^2}{\text{E}_\text{o}}$
$\Rightarrow \frac{\text{k}}{\text{E}_{\text{o}}}=\frac{3}{25}$
Similarly,
$\Rightarrow \text{t}_2=\frac{\text{k10}^2}{\text{E}_\text{o}}$
$\Rightarrow\text{t}_2=\frac{3}{25}\times10^2=12\text{ s}$

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MCQ 141 Mark

A photographic plate is placed directly in front of a small diffused source in the shape of a circular disc. It takes $12s$ to get a good exposure. If the source is rotated by $60^\circ$ about one of its diameters, the time needed to getthe same exposure will be:

A
$6s$
B
$12s$
✓
$24s$
D
$48s.$

Answer

Correct option: C.

$24s$

Here,
$\text{t}_1=12\text{s}$
$\theta_1=0^0$
$\theta_2=60^0$
$\text{t}_2=?$
Let the distance be $r.$
Let the incident luminosity be $E_o$.
We have,
$\text{E}_{\theta1}=\frac{\text{E}_\text{o}\cos\theta_1}{\text{r}^2}$
$\text{t}_1 \alpha\frac{1}{\text{E}\theta_1}$
$\Rightarrow\text{t}_1=\frac{\text{r}^2\text{k}}{\text{E}_\text{o}\cos\theta_1}$
$\Rightarrow12=\frac{\text{r}^2\text{k}}{\text{E}_\text{o}\cos0}$
$\Rightarrow \frac{\text{r}^2\text{k}}{\text{E}_\text{o}}=12$
Simiolarly,
$\text{t}_2=\frac{\text{r}^2\text{k}}{\text{E}_\text{o}\cos\theta_2}=\frac{12}{\cos(60^0)}$
$\Rightarrow\text{t}_2=12\times2=24\text{s}$

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Question 151 Mark

The one parameter that determines the brightness of a light source sensed by an eye is:

Energy of light entering the eye per second.
Wavelength of the light.
Total radiant flux entering the eye.
Total luminous flux entering the eye.

Answer

Total luminous flux entering the eye.

Explanation:
Total luminous flux is the total brightness producing capability of a radiating source. Or, it is the measurement of the total energy entering our eyes that produces the sensation of vision.

This cannot be the correct answer because all energies cannot be sensed by our eyes.
This cannot be the correct answer because all wavelengths do not produce any sensation in our eyes.
This cannot be the correct answer because all wavelengths contributing the radiant flux are not always visible to our eyes. There may be a large radiant flux yet there may not be any sensation of vision.

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Question 161 Mark

Light from a point source falls on a screen. If the separation between the source and the screen is increased by 1%, the illuminance will decrease (nearly) by:

0.5%
1%
2%
4%

Answer

2%

Explanation:
Illuminance is given by:
$\text{E}=\frac{\text{l}_{\text{o}}\cos\theta}{\text{r}^2}$
$\theta=0^0$
$\frac{\triangle\text{r}}{\text{r}}=1\%$
$\text{E}=\frac{\text{l}_{\text{o}}}{\text{r}^2}$
Differentiating,
$\text{dE}=-2\frac{\text{l}_{\text{o}}}{\text{r}^3}\text{dr}$
As approximation differentials are replaced by $\triangle$,
$\triangle\text{E}=-2\frac{\text{l}_\text{o}}{\text{r}^2}\triangle \text{r}$
$$$\Rightarrow\triangle\text{E}=-2\frac{\text{I}_\text{o}}{\text{r}^2}\Big(\frac{\triangle\text{r}}{\text{r}}\Big)$
$\Rightarrow\triangle\text{E}=-2\text{E}\Big(\frac{\triangle\text{r}}{\text{r}}\Big)$
$\Rightarrow\frac{\triangle{\text{E}}}{\text{E}}=-2\Big(\frac{\triangle{\text{r}}}{\text{r}}\Big)$
$\Rightarrow \frac{\triangle\text{E}}{\text{E}}=-2\times1\%=-2\%$
Since, negative sign implies decrease; hence, illuminance decreases by 2%.

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Question 171 Mark

Mark the correct options.

The luminous efficiency of a monochromatic source is always greater than that of a white light source of same power.
The luminous efficiency of a monochromatic source of wavelength 555nm is always greater than that of a white light source of same power.
The illuminating power of a monochromatic source of wavelength 555nm is always greater than that of a white light source of same power.
The illuminating power of a monochromatic source is always greater than that of a white light source of same power.

Answer

The luminous efficiency of a monochromatic source of wavelength 555nm is always greater than that of a white light source of same power.
The illuminating power of a monochromatic source of wavelength 555nm is always greater than that of a white light source of same power.

Explanation:

The luminous efficiency of a monochromatic source may be less than that of the white light if the former emits wavelength far away from 555nm.
Yes, it is true that our eyes mostly respond to colours close to the wavelength of 555nm and detect them bright. So, luminous efficiency is unity (highest).
It is true because white light distributes its energy amongst certain colours that our eyes cannot detect as brightly as they detect a 555nm light.
It is not necessarily true. If the monochromatic light radiates in a wavelength that is far away from 555nm, our eyes will not perceive it as bright. So, it will have lesser illuminating power.

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Question 181 Mark

The brightness producing capacity of a source:

Does not depend on its power.
Does not depend on the wavelength emitted.
Depends on its power.
Depends on the wavelength emitted.

Answer

Depends on its power.
Depends on the wavelength emitted.

Explanation:
Brightness depends upon how our eyes perceive light. Our eyes perceive yellow colour the most, so brightness depends upon the colour of the source. Now, colour is related to the wavelength of the source; so, brightness depends upon the wavelength as well.
Our eyes detect brightness by the amount of photons actually reaching our retinas. Again, the number of photons depends upon the power of the source. So, brightness depends upon the power of the source too.

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MCQ 191 Mark

A point source of light moves in a straight line parallel to a plane table. Consider a small portion of the table directly below the line of movement of the source. The illuminance at this portion varies with its distance $r$ from the source as:

A
$\text{I}\propto\frac{1}{\text{r}}$
B
$\text{I}\propto\frac{1}{\text{r}^2}$
✓
$\text{I}\propto \frac{1}{\text{r}^3}$
D
$\text{I}\propto\frac{1}{\text{r}^4}$

Answer

Correct option: C.

$\text{I}\propto \frac{1}{\text{r}^3}$

Let the distance between the parallel straight lines be $L$.
Angle with normal $=\theta$
We know,
$\text{I}=\frac{\text{I}_\text{o}\cos\theta}{\text{r}^2}$
From the above figure, we get
$\text{I}=\frac{\text{I}_\text{o}\sin(90^0-\alpha)}{\text{r}^2}$
$\Rightarrow\text{I}=\frac{\text{I}_\text{o}\sin\alpha}{\text{r}^2}$
$\Rightarrow\text{I}=\frac{\text{I}_\text{o}}{\text{r}^\text{2}}\Big(\frac{\text{L}_\text{o}}{\text{r}\text{}}\Big)$
$\Rightarrow\text{I}=\frac{I_\text{o}}{\text{r}^2}\Big(\frac{\text{L}}{\text{r}}\Big)$
$L =$ constant for parallel moving source
So $, I_o\ L =\text{K} \big($constant$\big)$
$\Rightarrow\text{I}=\frac{\text{k}}{\text{r}^3}$
$\Rightarrow\text{I}\alpha\frac{1}{\text{r}^3}$

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Question 201 Mark

Using figure find the relative luminosity of wavelength:

480nm
520nm
580nm
600nm

Answer

From the graph, we can find the following:

The relative luminosity of wavelength 480nm is 0.14.
The relative luminosity of wavelength 520nm is 0.68.
The relative luminosity of wavelength 580nm is 0.92.
The relative luminosity of wavelength 600nm is 0.66.

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Question 211 Mark

Light is incident normally on a small plane surface. If the surface is rotated by an angle of $30^\circ$ about the incident light, does the illuminance of the surface increase, decrease or remain same? Does your answer change if the light did not fall normally on the surface?

Answer

If the surface is rotated by $30^\circ ,$ the illuminance will decrease.
This is because illuminance depends upon the cosine of the normal angle.
Yes, if the light does not fall normally on the surface initially, it may increase or decrease depending upon the former's angle. If the $30^\circ$ rotation brings the table closer to the normal of the surface, the illuminance will increase; otherwise, it will decrease

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Question 221 Mark

The sun is less bright at morning and evening as compared to at noon although its distance from the observer is almost the same. Why?

Answer

During noon, the Sun's rays fall directly on the Earth's surface. The Sun being a yellow star, its rays also appear yellow at noon. Since yellow light has a high relative luminosity, it produces a high sensation of visibility in our eyes, thereby making the Sun appear brighter.
However, in the morning and evening, due to the slanting rays of the Sun on the Earth's surface, the smaller and middle range wavelength of the rays get scattered in the upper atmosphere. Thus, the sun appears orange during these two times. Since red/ orange light has a low relative luminosity, it produces a low sensation of visibility in our eyes, thereby making the Sun appear less brighter.

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