Class 10 Science (1Marks) Physics

This test is strictly based on the curriculum of SEE science class 10 for one marks questions. Your feedbacks are heartily welcome.

Science Practice Exam 10

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Electric Field

Here is a complete note on Electtric field.

Capacitors

Here is a complete note on Capacitors.

Force Note

Here is a complete note on chapter one force.

Question Electricity and magnetism

d) What type of combination of cells is shown in the given circuit? Calculate the power of the bulb if 0.2Acurrent flows through the curcuit.

 

Answer the  following questions:

a) What is ment by one kilowatt hour in electricity?
Ans: The electricity consumed by electric load of load of one kilowatt power in one hour is known as one kilowatt hour in electricity.

b) What is meant by electric current?
Ans: The flow of electron in a conducting wire is known as electric current.

c) What is meant by electric circuit?
Ans: The pathway of electricity containing combination of source, load and switch is known as electricity circuit.

d) What is meant by electricity consumption?
Ans: The amount of electricity which id converted to the other forms of energy  by using load is known as electricity consumption.

e) How is electricity consumption calcutated?
Ans: To claculate the electricity consumption the product of the number of load, power, and time is calculated:
Electricity consumption= P×N×T

f) What is meant by electric power?
Ans: The amount of electric of energy consumd by a load in one second is known as electric power.

g) Which instrument is used to measure the electrical power consumption?
Ans: Meter box is used to measure the electrical power consumption. It measures the power consumption in kilowatt hour.

h)  What is meant by galvanometer?
Ans: The electrical device which is used to measure small signals of current is known as galvanometer.

i) On which principle generator and dynamo depends?
Ans: The working principle of generator and dynamo depends on the principle of electromagnetic induction.

j) What are the types of generator?
Ans: Following are the types of generator:
i. AC generator
ii. DC generator, etc.

Give reasons:

a) The efficiency of a transformer is never 100%, why?
Ans: There is resistance in the coil of a transformer, so the efficiency of a transformer is never 100%.

b) Soft iron armature is used in electric bell, why?
Ans: The more the metal is soft the strength of magnet formed is more. So, soft iron armature is used in electric bell.

c) Filament lamp is filled with inert gases. Why?
Ans: In high temperature i.e. around 2900°C tungsten filament can get oxidised due to reaction with oxygen. So, to prevent oxidation and maintain the pressure, innert gas is filled in filament lamp.

d) Why is Nichrome used in heater?
Ans: Nichrome has high resistance and does not get oxidised even in high temperature like 900°C and can easily converts electric energy into heat energy. So, nichromee is used in heater.

e) Why should the coil of transformer be laminated?
Ans: When curent flows through the coil of transformer, eddy current is produced in the core. If the coil of transformer becomes a single piece, the amount of eddy current increases and the transformer gets heated and damaged. Output is also low. So, the  coil of transformer is laminated.

Differentiate between:

a) Phase wire and Neutral wire
Ans: Following are the differentiate between on phase and neutral wire:

Phase wire	Neutral wire
It is a red/ brown colored wire connected electrical device (load) to input current.	It is a black or blue colored wire through which current from load returns back.
We feel electric on touching it.	We feel electric on touching it.

Numerical problems:

i. Power consumption (pc): Power (P) × Hour(H)
=Watt × Hours
= KWH
=Unit

ii. Watt = Volt × Amper
Power = Volt (V) × current (Amper)
According to ohm’s law:
V = IR
PD = Current × Resistance
Volt     Amper      ohm (Ω)

a) An electric kettle rated 220 V and 2.2 KW works for 3 hours. Find the energy consumption and current drawn by it.
Given data,
PD = 220 volt
Power= 2.2 kw
Time = 3 hours
We know that,

Power consumption= power × hour
= 2.2kw × 3 hour
=6.6 KWH
= 6.6 Unit

Again, current drawn (I)= ?
P=VI
∴ I = P/V = 2.2 × 100W/ 220 = 2200/220
=10 AMP

∴The energy consumption is 6.6 unit and current drawn by it is 10 AMP.

b) If a 1000W heater is to be operated at a 220V mains, what must be the rating of the force to be used in the circuit?
Given data,
P = 1000 W
PD= 220V
Fuse (I)= ?
We know that,
P =VI
∴I = P/V= 1000/220
=50/11 = 4.5 = 5AMP

∴So the 5AMP is the rating of the force to be used in the circuit.

c) If two irons of 750W each are used 8hours a month. How much tariff should be paid? Cost of one unit of electricity is Rs 7.
Given data,
P= 750W
T=2 month
N= 8 hours
We know that,
Power consumption= 750 × 8 × 2
= 2 × 750 × 8/ 1000
= 12 KHW
=2 unit

∴ The cost of 1 unit= Rs. 7.00
∴The cost of 12 unit = 12×7 = Rs. 84.00

d) What type of combination of cells is shown in the given circuit? Calculate the power of the bulb if 0.2Acurrent flows through the curcuit.

Given data,
Amp= 0.2
Volt = 6
We know that,
Watt= Volt × Amp
= 6×0.2
=1.2 watt

e) In a transformer, the number of turns in primary coil is 100. If the primary and secondary voltage are 220V and 110V respecctively, calculate the no of turns in the secondary coil.
Given data,
Np= 100
Vp= 220 volt
Vs= 110 volt
Ns=?
we know that,

Question Light

 

Answer these following questions

1. What is the lens? Mention its types.
   Ans: The lens is the medium of refraction of light.
   Its types are:
   a) Convex lens
   b) Cocave lens, etc.
2. Define the following:
   a) Centre of curvature: The center of the sphere of which the lens part is called the center of curvature.b) Principal axis: An imaginary line which joins two centers of curvatures (C1 and C2) is called Principal axis.c) Optical center: The center of a lens which is equidistant from all points of its surface is called Optical center.d) Principal focus: Principal focus is a point on the principal axis where the rays of light parallel to the principal axis coverage after refraction through a convex lens or appear to diverge after refraction through a concave lens.e) Focal length: The distance between the optical center and the principal focus is called Focal length.
3. What is magnification? Mention the formula for finding the magnification of an object.
   Ans: Magnification is a factor which determines image is how many times larger or smaller than an object.
   The formula for finding the magnification of an object
   Magnification (m)= size of image/size of an object
   = Image distance (v)/ Object distance(0)
4. A convex lens forms a real and magnified image where is the object placed?
   Ans: A convex lens forms a real and magnified image when it is the object placed between 2F and F.
5. What is the power of the lens? Mention its formula.
   Ans: The conversing or diverging capacity of a lens is called Power of lens.
   Power of lens (P)= 1/ focal length (m)
6. Mention the lens formula.
   Ans: Power of lens (P)= 1/ focal length (m)
7. Mention the uses of the lens.
   Ans: The uses of the lens are:
   a) It is used in a camera.
   b) It is used in remedy defect of eyesight, etc.
8. What are optical instruments?
   Ans: The instruments which produce the image of an object are called optical instruments.
9. How does pupil control the amount of light entering the eyes?
   Ans: Pupil controls the amount of light entering the eye by changing the size of a pupil by iris.
10. What is accommodation?
    Ans: The ability of the eye to focus the image of an object on the retina by changing the focal length of its lens is called Accommodation.
11. What is Near point?
    Ans: The nearest point up to which a normal eye can see clearly is called Near point.
12. Define Far point.
    Ans: The farthest point up to which a normal eye can see clearly is called Far point.
13. What is Range of vision?
    Ans: The distance between near point and far point is called Range of vision.
14. Define Defect of vision.
    Ans: When an eye cannot see the object lying at the range of vision is called Defect of vision.
15. Ram cannot see the letter written on the blackboard. What is his defect of vision? How can it be corrected?
    Ans: Ram cannot see the letter written on the blackboard because he is suffering from Myopia and he needs to wear goggles rear with the concave lens.

Give reasons:

1. Shortsightedness can be removed by using the concave lens.
   Ans: Shortsightedness can be removed by using the concave lens because it concentrates light coming from the distant object of near point (N).
2. A concave lens is called a diverging lens.
   Ans: A concave lens is called a diverging lens because of it diverse parallel light from a point.
3. A convex lens is called a converging lens.
   Ans: A convex lens is called a converging lens because it converses parallel light at a point (focus).
4. Power of a convex lens is positive.
   Ans: Power of a convex lens is positive because the real distance between the optical center and the real image is taken as positive.

Numerical Problems

a) Find the power of a lens of focal length 10 cm.
Given data,
f= 10 cm
= 10/100 cm
we know that,
p=1/f(m)
=1/10/100
= 100/10
= 10 D
∴ Power of focal length is 10 D.
b) What is the focal length of a lens having the power 5D?
Given data,
p= 5D
f=?
we know that,
p=1/f(m)
f(m)= 1/p = 1/5 -= 0.2 m.
∴ The focal length of a lens is 0.2m.
c) An object is placed at a distance of 60cm in front of a convex lens of a focal length 20cm. where is the image formed? Mention magnification of the image and power of the lens.
Given data,
u=60cm
f=20cm
v=?
m=?
p=?
we know that,
1/f = 1/u + 1/v
1/20 = 1/60 + 1/v
1/20 + 1/60 = 1/v
3+1/60 = 1/v
or, 1/v= 4/60 =1/15
∴v= 15 cm
Now,
Magnification (m)= Image distance(v) / Object distance(u)
=15/60
=1/4
=0.25
Again, p=1/f(m)
=1/20
=1/20/100
=100/20
=5D
∴The magnification of the image is o.25 and power lens is 5D.
d) An object is placed at the distance of 40 cm from the concave lens of focal length 20cm. Find the image distance.
Given data,
u= -40
f= -20cm
we know that,
1/f= 1/u+1/v
1/-20= 1/-40 = 1/v
1-1/40= 1/v
-1/40= 1/v
-v=40
∴v= -40
∴The image distance is -40cm.
e) A student wears spectacles of power -2d. What types of defect of the eye does the student have? Find the focal length of the lens used?
Given data,
p=-2D
f=?
we know that,
p= 1/f(m)
or, -2= 1/f(m)
or, f(m)= 1/-2
∴f(m)= -50cm

∴The focal length of the lens used is -50cm.

Energy Question Answer

 

1. What is the primary and secondary source of energy?
   Ans: The source of energy which can be used in the same form in which they present naturally are called the primary source of energy.
   The source of energy which is obtained from the primary source is called the secondary source of energy.
2. What is renewable energy? Give some examples.
   Ans: The source of energy which can be obtained again and again when one exhausted is called the Renewable source of energy.
   Examples: Biogas, Nuclear fuel, etc.
3. What is the Nonrenewable source of energy? Give some examples.
   Ans: The source of energy which can’t be obtained when exhausted is called the Non-Renewable source of energy.
   Examples: Fossil fuel, Petrochemical, etc.
4. What is a fossil fuel? Give some examples.
   Ans: The source of energy which can be obtained by dead decayed plants and animals of millions of years ago is called a fossil fuel.
   Example: Coal, petrol, etc.
5. What is hydro-power? Mention its advantages.
   Ans: The power produced by the power of water is called Hydropower.
   Its advantages are:
   a) It is pollution free.
   b) It is a renewable source of energy.
   c) It transforms from one form to another.
   d) It operates on modern devices, etc.
6. What is the energy crisis? What is the various way to solve the problem of the energy crisis?
   Ans: The problem created by the exhaustion of fossil fuel reserves due to excessive use is called energy crisis.
   The various way to solve the problem of the energy crisis is given below:
   a)By avoiding unnecessary use of electricity.
   b)By conserving the existing source of energy.
   c)By producing and using more hydro-electricity.
   d)By developing the alternating source of energy, etc.
7. What is the alternate source of energy? What is the various alternate source of energy?
   Ans: The substitute for non-renewable energy is called the Alternate source of energy. The alternate source of energy is the renewable source of energy.
   The various source of alternate energy is given below:
   a)Biogas,
   b)Solar energy, etc.
8. What is Biogas? How is it obtained?
   Ans: Biogas is produced from organic matter which can be used as an energy source. It can be obtained from animal dung.
9. What is the advantage of biogas?
   Ans: The advantage of biogas are given below:
   a)It doesn’t produce smoke.
   b) It can be used for cooking food.
   c) It is a renewable source of energy, etc.
10. What is nuclear fuel? What are the various types of nuclear reactions?
    Ans: The elements which produce a large amount of energy as a result of nuclear reactions are nuclear fuels.
    The various types of nuclear reactions are given below:
    a)Nuclear fission
    b)Nuclear fusion, etc.
11. Mention advantage of nuclear energy.
    Ans: Advantage of nuclear energy are given below:
    a) A small amount of nuclear energy produces a huge amount of energy.
    b) It is used by few, so it can be alternate energy, etc.
12. What is wind energy? Mention its advantages.
    Ans: The energy which contains wind at high speed is called wind energy.
    Its advantages are:
    a)It is free of cost.
    b) It is a renewable source of energy.
    c) It can be used to generate electricity.
    d) It provides energy without depleting energy, etc.
13. What is geothermal energy? Mention its advantages.
    Ans: The energy obtained by heat energy available inside the earth is called Geothermal Energy.
    Its advantages are given below:
    a) It supplies energy without pollution.
    b) It is a perpetual source of energy.
    c) It can generate electricity, etc.
14. What is solar energy? Write its uses?
    Ans: The energy obtained from the sun is called Solar energy.
    Its uses are:
    a) For drying clothes.
    b) For preserving fruits.
    c) For providing heat and light to the earth.
    d) For obtaining heat and light energy, etc.
15. What are the two natural processes of obtaining geothermal energy?
    Ans: The two natural process of obtaining geothermal energy are given below:
    a)Volcanic eruption.
    b)Hot water spring, etc.

Give reasons:

1. Wind energy is the outcome of solar energy.
   Ans: Wind energy is the outcome of solar energy because the heat of the sun causes the air of one planet to be heated. The hot air being lighter rises up and a vacant place is created. The cooler air from other places moves towards such vacant with great speed.
2. Biofuel and fossil fuel is the outcome of solar energy.
   Ans: Biofuel and fossil fuel is the outcome of solar energy because plants convert solar energy into chemical energy and store it in their body. The biomass of plants and animals are used as biofuel. When plants and animals of million years ago died and got buried under the soil, they were converted into fossils.
3. Hydroelectricity is the outcome of solar energy.
   Ans: Hydroelectricity is the outcome of solar energy because Iceberg is melt due to the heat of solar energy then it follows in high that is used to produce electricity.
4. Mineral oil is a fossil fuel.
   Ans: Mineral oil is a fossil fuel because mineral oil is formed from dead bodies buried millions of years ago.
5. Biogas is the best energy source in Nepal.
   Ans: Biogas is the best energy source in Nepal because Nepal is based on agriculture, so dung produce from animals is used to produce biogas.
6. Production of hydroelectricity should be encouraged.
   Ans: Production of hydroelectricity should be encouraged because the production cost of hydroelectricity is high at the beginning, after that it will be run for a long time and produces heat.

Electromagnetic Induction

Electric Circuit

Interference

a InterferenceCoherent Sources and Sustained Interferencenterference is a natural phenomenon that happens at every place and at every moment. Yet we don’t see interference patterns everywhere. Interference is the phenomenon in which two waves superpose to form the resultant wave of the lower, higher or same amplitude. The most commonly seen interference is the optical interference or light interference. This is because light waves are randomly generated every which way by most sources. This means that light waves coming out of a source do not have a constant amplitude, frequency or phase. The most common example of interference of light is the soap bubble which reflects wide colours when illuminated by a light source.Coherent SourcesTwo sources are said to be coherent when the waves emitted from them have the same frequency and constant phase difference.Interference from such waves happen all the time, the randomly phased light waves constantly produce bright and dark fringes at every point. But, we cannot see them since they occur randomly. A point that has a dark fringe at one moment may have a bright fringe at the next moment. This cancels out the effect of the interference effect, and we see only an average brightness value. The interference is not said to be sustained since we cannot observe it.Characteristics of Coherent SourcesCoherent sources have the following characteristics:• The waves generated have a constant phase difference• The waves are of a single frequencyCoherent Source Example• Laser light is an example of coherent source of light. The light emitted by the laser light has the same frequency and phase.• Sound waves are another example of coherent sources. The electrical signals from the sound waves travel with the same frequency and phase.Types of InterferenceInterference of light waves can be either constructive interference or destructive interference. Constructive interference: Constructive interference takes place when the crest of one wave falls on the crest of another wave such that the amplitude is maximum. These waves will have the same displacement and are in the same phase. for constructive interference, the path difference between the waves from two sources is path difference, QX–PX=n𝜆$$\text{path difference,}QX–PX=n𝜆$$ where n is the integer number.Destructive interference: In destructive interference the crest of one wave falls on the trough of another wave such that the amplitude is minimum. The displacement and phase of these waves are not the same. For destructive interference, we have path difference, QY–PY=(n+1

2)𝜆$$\text{path difference,}QY–PY=(n+\frac{1}{2})𝜆$$ where n is the integer number.Conditions for Interference of Light Waves For sustained interference of light to occur, the following conditions must be met:• Coherent sources of light are needed.• Amplitudes and intensities must be nearly equal to produce sufficient contrast between maxima and minima.• The source must be small enough that it can be considered a point source of light.• The interfering sources must be near enough to produce wide fringes.• The source and screen must be far enough to produce wide fringes.• The sources must emit light in the same state of polarization.• The sources must be monochromatic. Path Difference and Phase DifferenceWhen a wave passes through a medium, the particles of the medium vibrate. When the particles completer one to and fro motion, the wave advances by a distance equal to its wavelength λ. For a complete wave, the wavelength varies in λ and the phase is changed through 2𝜋${\displaystyle 2𝜋}$.. Let there be two waves with a path difference of λ. Then, the phase difference them will be 2𝜋${\displaystyle 2𝜋}$. If the path difference is x, then path difference = 2𝜋

𝜆 × x${\displaystyle =\frac{2𝜋}{𝜆}\times x}$ hence, phase difference=2𝜋

𝜆×path difference$$\text{hence, phase difference}=\frac{2𝜋}{𝜆}\times \text{path difference}$$Optical PathThe product of the distance travelled by the light in a medium and the refractive index of that medium is called the optical path. If d be the distance travelled by the light in a medium of refractive index µ, then by definition, Optical Path=𝜇d$$\text{Optical Path}=𝜇d$$ Young’s Double Slit Experiment

Simulation

S is a narrow vertical slit (of width about 1 mm) illuminated by a monochromatic source of light. At a suitable distance (about 10 cm ) from S, there are two fine slits S1${\displaystyle S_1}$ and S2${\displaystyle S_2}$ about 0.5 mm apart at equidistant from S. when a screen is placed at a larger (about 2m) from the slits S1${\displaystyle S_1}$ and S2${\displaystyle S_2}$, alternate bright and dark bands appear on the screen. The appearance of bright and dark bands are called the fringes.Theory of Interference of Light Suppose S1${\displaystyle S_1}$ and S2${\displaystyle S_2}$ be two fine slits at a small distance d apart in the figure. Let slits are illuminated by monochromatic light from a strong source S of wavelength λ and MN is a screen at a distance D from the double slits. The two waves starting from S1${\displaystyle S_1}$ and S2${\displaystyle S_2}$ superimpose upon each other resulting an interference pattern on the screen placed parallel to the double slit as in the figure.

Theory of interference fringesLet O be the centre between the slits S1${\displaystyle S_1}$ and S2${\displaystyle S_2}$. Draw S1P, S2P ${\displaystyle S_{1}P,S_{2}P}$and OC perpendicular to MN. The intensity of light at a point on the screen will depend upon the path difference between the two waves arriving at the point. The point C on the screen lies on the perpendicular bisector of S1${\displaystyle S_1}$ and S2${\displaystyle S_2}$. Therefore, the path difference between two waves reaching C is zero and hence, they are in phase. So, the point C is the position of maximum intensity. It is called central maximum. Consider a point P at a distance x from C. The path difference between two waves arriving at P is given by path difference=S2P–S1P$$\text{path difference}=S_{2}P–S_{1}P$$ From the geometry in figure, it is found that

	PQ=x-d 2;PR=x+d 2
	And(S2P)2–(S1P)2 =[D2+(x+d 2)2]-[D2+(x-d 2)2]=2xd
	or, (S2P–S1P)(S 2P+S1P)=2xd
	or, (S2P–S1P)=2xd BP+AP
	In practice, point P lies very close to C.
	So S2P≈S1P≈D
	S2P+S1P=D+D=2D
	Path difference=S2P–S1P=2xd 2D=xd D

$$\begin{array}{c}PQ=x-\frac{d}{2};PR=x+\frac{d}{2}\\ \text{And}(S_{2}P{)}^2–(S_{1}P{)}^2=[D^2+(x+\frac{d}{2}{)}^2]-[D^2+(x-\frac{d}{2}{)}^2]=2xd\\ \text{or,}(S_{2}P–S_{1}P)(S_{2}P+S_{1}P)=2xd\\ \text{or,}(S_{2}P–S_{1}P)=\frac{2xd}{BP+AP}\\ \text{In practice, point P lies very close to C.}\\ \text{So}{S}_{2}P\approx S_{1}P\approx D\\ S_{2}P+S_{1}P=D+D=2D\\ \text{Path difference}=S_{2}P–S_{1}P=\frac{2xd}{2D}=\frac{xd}{D}\end{array}$$ The waves from S1${\displaystyle S_1}$ and S2${\displaystyle S_2}$ arriving at a point on the screen will interfere constructively or destructively depending upon this path difference. The phase difference for this path difference is given by Phase difference, 𝜙=2𝜋

𝜆(xd

D )$$\text{Phase difference,}𝜙=\frac{2𝜋}{𝜆}\left(\frac{xd}{D}\right)$$ Bright fringesIf the path difference is an integral is an integral multiple of wavelength λ, then point P is bright. Therefore, for bright fringes

	xd D	=n𝜆
	x	=n𝜆D d…(i)

$$\begin{array}{rl}\frac{xd}{D}& =n𝜆\\ x& =n𝜆\frac{D}{d}\dots (i)\end{array}$$ where n = 0, 1, 2, 3, … The distance of the various bright fringes from the central maximum at C can be found as follows:

	For n=0,x0=0… central bright fringes
	For n=1,x1=𝜆D d…first bright fringes
	For n=2,x2=2𝜆D d…second bright fringes
	For n=n,xn=n𝜆D d…nth bright fringes

$$\begin{array}{c}\text{For}n=0,x_0=0\dots \text{central bright fringes}\\ \text{For}n=1,x_1=\frac{𝜆D}{d}\dots \text{first bright fringes}\\ \text{For}n=2,x_2=\frac{2𝜆D}{d}\dots \text{second bright fringes}\\ \text{For}n=n,x_n=n𝜆\frac{D}{d}\dots n^{th}\text{ bright fringes}\end{array}$$The distance between any two consecutive bright fringes is called fringe width, denoted by β.

	Fringe width, 𝛽=x2–x1=2𝜆D d-𝜆D d=𝜆D d
	∴𝛽=𝜆D d…(ii)

$$\begin{array}{c}\text{Fringe width,}𝛽=x_2–x_1=\frac{2𝜆D}{d}-\frac{𝜆D}{d}=\frac{𝜆D}{d}\\ \therefore 𝛽=\frac{𝜆D}{d}\dots (ii)\end{array}$$Dark fringesIf the path difference is an odd integral multiple of half wavelength λ, then point P is dark. Therefore, for dark fringes;

	xd D	=(2n -1)𝜆 2 where n=1,2,3,…
	or, x	=(2n-1)𝜆D 2d…(iii)

$$\begin{array}{rl}\frac{xd}{D}& =(2n-1)\frac{𝜆}{2}\text{where}n=1,2,3,\dots \\ \text{or,}x& =(2n-1)\frac{𝜆D}{2d}\dots (iii)\end{array}$$Equation (iii) gives the distance of the dark fringes from point C. the distance of the various dark fringes from point C can be calculated as below:

	For n=1,x1=𝜆D 2d…first dark fringe
	For n=2,x2= 3𝜆D 2d…second dark fringes
	For n=n,xn=(2n-1)𝜆𝜆D 2d…nth dark fringes

$$\begin{array}{c}\text{For}n=1,x_1=\frac{𝜆D}{2d}\dots \text{first dark fringe}\\ \text{For}n=2,x_2=\frac{3𝜆D}{2d}\dots \text{second dark fringes}\\ \text{For}n=n,x_n=(2n-1)𝜆\frac{𝜆D}{2d}\dots n^{th}\text{ dark fringes}\end{array}$$The distance between any two consecutive dark fringes is called fringe width β, given as

	Fringe width, 𝛽=x2–x1=3𝜆D 2d-𝜆D 2d=𝜆D d
	∴𝛽=𝜆D d…(iv)

$$\begin{array}{c}\text{Fringe width,}𝛽=x_2–x_1=\frac{3𝜆D}{2d}-\frac{𝜆D}{2d}=\frac{𝜆D}{d}\\ \therefore 𝛽=\frac{𝜆D}{d}\dots (iv)\end{array}$$from equation (ii) and (iv), it is clear that width of the bright is equal to the width of the dark fringe. From these two equations it is clear that fringe width increases as the1. Wavelength increases.2. Distance D of the screen from the sources increases3. Distance between the sources decreases.