Students must start practicing the questions from RBSE 12th Physics Model Papers Set 1 with Answers in English Medium provided here.

## RBSE Class 12 Physics Model Paper Set 1 with Answers in English

Time . 2 Hours 45 Min.

Max Marks: 56

General Instruction to the Examinees:

- Candidate must write first his/her Roll No. on the question paper compulsorily.
- All the questions are compulsory.
- Write the answer to each question in the given answer book only.
- For questions having more than one part the answers to those parts are to be written together in continuity.

Section-A

Multiple Choice Questions

Question 1.

Write the Correct Answer from multiple choice question 1 (i to ix) and write in given answer book:

(i) Gauss’s law should be invalid if: [1]

(a) There were magnetic monopoles

(b) The inverse square law were not exactly true

(c) The velocity of light were not a universal constant

(d) None of these

Answer:

(b) The inverse square law were not exactly true

(ii) An electric dipole is placed at the centre of a hollow conducting sphere. Which of the following is correct? [1]

(a) Electric field is zero at every point of the sphere.

(b) Electric field is not zero everywhere on the sphere.

(c) The flux of electric field is not zero through the spehre.

(d) All of these

Answer:

(b) Electric field is not zero everywhere on the sphere.

(iii) Combine three resistors 5 Ω, 4.5 Ω and 3 Ω in such a way that the total resistance of this combination is maximum. [1]

(a) 14.5 Ω

(b) 13.5 Ω

(c) 12.5 Ω

(d) 16.5 Ω

Answer:

(c) 12.5 Ω

(iv) The cyclotron frequency v is given by: [1]

(a) \(\frac{q B}{2 \pi m}\)

(b) \(\frac{m B}{2 \pi q}\)

(c) \(\frac{2 \pi m}{q B}\)

(d) \(\frac{2 \pi B}{q m}\)

Answer:

(a) \(\frac{q B}{2 \pi m}\)

(v) The total charge induced in a conducting loop when it is moved in magnetic field depends on: 1

(a) The rate of change of magnetic flux

(b) Initial magnetic flux only

(c) The total change in magnetic flux and resistance

(d) Final magnetic flux only

Answer:

(c) The total change in magnetic flux and resistance

(vi) Which of the following figure represents the variation of particle momentum and the associated de-Broglie wavelength? [1]

Answer:

(vii) Equivalent energy of mass equal Lo 1 amu is: [1]

(a) 931 KeV

(b) 931 eV

(c) 931 MeV

(d) 9.31 MeV

Answer:

(c) 931 MeV

(viii) In the energy bond diagram of a material shown below, the open circles and filled circles denote holes and electrons respectively. The material is: [1]

(a) A p-type semiconductor

(b) A insulator

(c) A metal

(d) An n-type semiconductor

Answer:

(a) A p-type semiconductor

(ix) The reverse biasing in a PATjunction diode: [1]

(a) Decreases the potential barrier

(b) Increases the potential barrier

(c) Increases the number of minority charge carriers

(d) Increases the number of majority charge carriers

Answer:

(b) Increases the potential barrier

Question 2.

Fill in the blanks:

(i) …………………….. is the factor which decided the direction of flow of electric charge. [1]

Answer:

Electric Potential

(ii) The current flowing normally per unit area of cross-section is known as …………………… . [1]

Answer:

Current density

(iii) …………………….. is a device which converts mechanical energy into electrical energy. [1]

Answer:

dynamo

(iv) The process of converting a.c. into d.c. is called ………………………… [1]

Answer:

rectification

Question 3.

Give the answer of the following questions in one line (i to viii).

(i) How is the drift velocity in a conductor affected with the rise in temperature? [1]

Answer:

The average drift velocity,

υ_{e} = \(\frac{e E}{m}\)τ

where, τ = relaxation time.

The relaxation time is directly proportional to the temperature of conductor, i.e.,

τ ∝ T

∴ υ_{d} ∝ T

So, the drift velocity increases with rise in temperature.

(ii) State two properties of the material of the wire used for suspension of the coil in a moving coil galvanometer . [1]

Answer:

- Non-Brittle, conductor.
- Restoring torque per unit twist should be small.

(iii) Define the term self inductance of a coil. Write its SI unit. [1]

Answer:

Self-inductance of a coil is equal to the total magnetic flux linked with , the coil, when unit current passes through it.

Also, self-inductance of a coil, is equal to the emf induced in coil, when rate of change of current in coil is 1 A/s.

SI unit of self-inductance is 1 Henry (H)

1 H = 1 V-s/A

(iv) The stopping potential in an experiment on photoelectric effect is 2V. What is the maximum kinetic energy of the photoelectrons emitted? [1]

Answer:

Given, stopping potential in an experiment on photoelectric effect = 2V

Maximum kinetic energy

KE_{max} = eV_{0}

e (2V) = 2eV

(v) To observe photoelectric effect the frequency of incident light should be more than which frequency? [1]

Answer:

To observe photo-electric effect the frequency of incident light should be greater than the threshold frequency of the material.

(vi) Why is it found experimentally difficult to detect neutrinos in nuclear β-decay? [1]

Answer:

Neutrinos are difficult to detect because they are massless, have po charge, do not interact with nucleons and can penetrate large quantity of matter (even the earth) without any interaction.

(vii) State two characteristic properties of nuclear force. [1]

Answer:

- Nuclear forces are the strongest force in nature.
- They are saturated forces

(viii) Write the truth table for a Not gate connected as shown in the figure. [1]

Answer:

Truth Table

Question 4.

Explain using suitable diagrams, the behaviour of a dielectric in the presence of external electric field. Define the terms polarisation of a dielectric and write its relation with susceptibility. [1½]

Answer:

Dielectrics are non-conducting substances, i.e., they have no charge carriers. Thus, in a dielectric, free movement of charges is not possible. When a dielectric is placed in an external electric field, the molecules are re-oriented and thus induces a net dipole moment in the dielectric. This produces an electric field.

However, the opposing field is so induced that does not exactly cancel the external field. It only reduces it. Both polar and non-polar dielectric develop net dipole moment in the presence of an external field. The dipole moment per unit volume is called polarisation and is denoted by P for linear isotropic dielectrics.

Question 5.

Figures (a) and (b) show the field lines of a positive and negative point charge respectively. [1½]

(a) Give the signs of the potential difference (V_{P} -V_{Q} and (V_{B} – V_{A}).

(b) Give the sign of the potential energy difference of small negative charge between the points Q and P; A and B.

(c) Give the sign of the work done by the field in moving a small positive

charge from QtoP. 1 %

Answer:

(a) V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}\) ∴ V ∝ \(\frac{1}{r}\)

Therefore V_{P} > V_{Q} =>(V_{P} – V_{P})is positive.

And V_{B} is more positive than V_{A} i.e., V_{B} is less negative than V_{A}. Therefore (V_{B} – V_{A}) is positive.

(b) A small negative charge will be attracted towards positive charge. The negative charge moves from higher potential energy to lower potential energy. Therefore the sign of potential energy difference of a small negative charge between Q and P is positive.

(c) In moving a small positive charge from Q to P, work has to be done by an external agency against the electric field. Therefore, work done by the field is negative.

Question 6.

Using the concept of free electrons in a conductor, derive the expression for the conductivity of a wire in terms of number density and relaxation time. Hence, obtain the relation between current density and the applied electric field E. [1½]

Answer:

Conductivity

σ = \(\frac{1}{\rho}\)

We know that, drift velocity is given by

υ_{d} = \(\frac{e E \tau}{m}\) …………..(i)

where, e = electric charge,

E = applied electric field,

τ = relaxation time and m = mass of electron.

But E = \(\frac{V}{l}\) (i.e., potential gradient)

∴ υ_{d} = \(\left(\frac{e \tau}{m}\right)\left(\frac{V}{l}\right)\) …………………. (ii)

From the relation between current and drift velocity,

I = neAυ_{d} ………………. (iii)

(where, n = number of density of electrons)

Putting the value of Eq. (ii) in Eq. (iii), we get

I = neA(\(\frac{e \tau V}{m l}\)) or I = (\(\frac{n e^{2} A \tau}{m l}\)) V

But according to Ohm’s law,

V= IR ………….. (v)

R = (\(\frac{m}{m e^{2} \tau}\))\(\frac{l}{A}\) …………….. (vi)

Also R = ρ\(\frac{l}{A}\) …………….. (vii)

ρ = \(\frac{m}{n e^{2} \tau}\) = resistivity of conductor.

∴ Conductivity, σ = \(\frac{1}{\rho}=\frac{n e^{2} \tau}{m}\)

Now, we know that, current density,

J = \(\frac{I}{A}\)

or J = \(\frac{n e A v_{d}}{A}\) = neυ_{d} = )(\(\frac{n e^{2} \tau}{m}\))E

υ_{d} = \(\frac{e E \tau}{m}\)

∴ J = σE σ = \(\frac{n e^{2} \tau}{m}\)

Question 7.

In a metre-bridge the null point is found to be at a distance of 33.7 cm from A. If now a resistance of 12 Ω is connected in parallel with S, the null point occurs at 51.9 cm. Determine the resistance of R and S. [1½]

Answer:

In figure (A) \(\frac{R}{S}=\frac{A D}{D C}\)

\(\frac{33.7 \mathrm{~cm}}{66.3 \mathrm{~cm}}=\frac{33.7}{66.3}\)

In figure (B) when S is shunted by 12 Ω, it is changed to S’, where

S’= 12S/(12 + S)

Clearly \(\frac{R}{S^{\prime}}\) = \(\frac{51.9 \mathrm{~cm}}{48.1 \mathrm{~cm}}\)

or \(\frac{R(12+S)}{12 S}\) = \(\frac{51.9 \mathrm{~cm}}{48.1 \mathrm{~cm}}\)

or \(\frac{12+S}{12}=\frac{51.9}{48.1} \times \frac{S}{R}\)

= \(\frac{51.9}{48.1} \times \frac{66.3}{33.7}\) = 2.123

or \(\frac{S}{12}\) + 1 = 2.123

or S =(2.123 – 1)12Ω = 13.5Ω

Also, R = \(\frac{33.7}{66.3}\) S

= (\(\frac{33.7}{66.3}\))(13.5Ω) = 6.9Ω

Question 8.

In a circuit, two coils of self inductance L_{1} and L_{2} are combined in series, the mutual inductance between them is M. Determine the equivalent self inductance of the combination. [1½]

Answer:

According to question, the two coils of self inductance L_{1} and L_{2}; and mutual inductance M are combined in series. Let the induced emf in the two coils be ε_{1} and ε_{2} and the combined emf is e, then

e = ε_{1} + ε_{2} …………. (1)

If the rate of change of current in the two coils is \(\frac{d I}{d t}\), then

ε_{1} = -L_{1}\(\frac{d i}{d t}-\frac{M d I}{d t}\)

and, ε_{2} = -L_{2}\(\frac{d I}{d t}-\frac{M d I}{d t}\)

If the equivalent self inductance of the combination is L, then ε = -L \(\frac{d I}{d t}\)

∴ From eqn. (1),

–\(\frac{L d I}{d t}\) = -L_{1} \(\frac{d I}{d t}-\frac{M d I}{d t}\) -L_{2}\(\frac{d I}{d t}\)-M\(\frac{d I}{d t}\)

or, L = L_{1} + M + L_{2} + M

or, L = L_{1} + L_{2} + 2M

Question 9.

State Lenz’s law. A metallic rod held horizontally along east-west direction, is allowed to fall under gravity. Will there be an emf induced at its ends? Justify your answer. [1½]

Answer:

Lenz’s law states that the direction of induced emf of induced current in a circuit is such that it opposes the cause or the change that produces it. Yes, emf will be induced in the rod, as there is change in magnetic flux.

When a metallic rod held horizontally along East-West direction, is allowed to fall freely under gravity, i.e., fall from North to South, the intensity of magnetic lines of the earth’s magnetic field changes through it, i.e., the magnetic flux changes and hence the induced emf in it.

Question 10.

A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to? How is such a defect of vision corrected? [1½]

Answer:

This defect is called ‘astigmatism’. It arises because the curvature of the cornia plus eye lens refracting system is not the same in different planes. The eye lens is usually spherical, i.e., has the same curvature in different planes but the cotnia is not spherical in case of an astigmatic eye. In the present case, the curvature is the vertical plane is enough, so shop images of vertical lines can be formed on the retina. But the curvature is insufficient in the horizontal plane, so horizontal lines appear blurred. The defect can be corrected by using a cylindrical lens with its axis along the vertical. Clearly, parallel rays in the vertical plane will suffer no extra refraction, but those in the horizontal plane can get the required extra convergence due to refraction by the curved surface of the cylindrical lens if the curvature of the cylindrical surface is chosen appropriately.

Question 11.

Use the mirror equation to deduce that a convex mirror always produces a virtual image independent of the location of the object. [1½]

Answer:

From mirror formula,

\(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

or \(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{u-f}{u f}\)

∴ υ = \(\frac{u f}{(u-f)}\)

For concave mirror u and f both are negative, therefore

υ = \(\frac{(-u)(-f)}{(-u)-(-f)}=\frac{u f}{f-u}\)

Given: f < u < 2f

∴ (f – u) < 0 or (u – f) > 0

∴ υ = \(\frac{u f}{-(u-f)}\)

For convex f is positive and u is negative.

∴ υ = \(\frac{(-u) f}{-u-f}=\frac{u f}{u+f}\)

or υ = \(\frac{u f}{u+f}\)

It is clear that the value of v is positive, hence the image will form behind the mirror and it will be virtual.

Question 12.

Find the radius of curvature of the convex surface of a plano-convex lens, whose focal length is 0.3 m and the refractive index of the material of the lens is 1.5. [1½]

Answer:

For a plano-convex lens, R_{1} = ∞

R_{2} = – R f = 0.3 m = 30 cm

n = 1.5 ⇒ R = ?

Lens Maker’ formula,

\(\frac{1}{f}\) = (n – 1) (\(\frac{1}{R_{1}}-\frac{1}{R_{2}}\))

⇒ \(\frac{1}{30}\) = (n – 1) (\(\left.\frac{1}{\infty}-\frac{1}{-R}\right)\))

= (\(\frac{1}{30}=\frac{(1.5-1)}{R}\)) ⇒ R = 15 cm

Question 13.

What is mirage? [1½]

Answer:

Mirage: During summer, the air near the Earth’s surface becomes hot in comparison to upper layers of air. Hence, it becomes rarer and the density of air layers goes on increasing with increasing height. In such a situation, as the rays of light starting from a top of tree, move towards the Earth, they go on shifting away from the normal due to their movement from denser to rarer medium. During this process, in a particular situation, the rays undergoes total internal reflection and hence they appear to come from a point lower to the Earth’s surface. It means that the image of tree is formed and hence it appears, the tree is standing near the tank with its shade in the water of the tank. In this suspicion, when thirty animal goes there in search of water, it does not get water there. This is why this event is known as “Desert Mirage”.

Question 14.

Define activity of a radioactive material and write its SI unit. [1½]

Answer:

Activity: Sometimes when we are interested to determine the number of decayed nuclei per unit time rather than determining the number of active nuclei Nina sample (because it is more quantity to measure). It is called activity of the sample.

Activity, R = \(\left|\frac{d N}{d t}\right|\) …………… (4)

is called the initial activity. The S.I. unit of activity is Becquerel

Question 15.

A radioactive isotope has a half-life of 10 yr. How long will it take for the activity to reduce to 3.125? [1½]

Answer:

After n half lives, activity of sample is R = R_{0}(\(\frac{1}{2}\))^{n}

where, R_{0} = initial activity.

Given, T = 10 yr, R=3.125 R_{0}, n = \(\frac{t}{T}\)

where t = instantaneous time

We have, \(\frac{R}{R_{0}}\) = (\(\frac{1}{2}\)) ^{t/T}

∴ \(\frac{3.125}{100} \frac{R}{R_{0}}\) = (\(\frac{1}{2}\)) ^{t/100}

⇒ \(\frac{1}{32}\) = (\(\frac{1}{2}\)) ^{5} = (\(\frac{1}{2}\)) ^{t/10} ⇒ 5 = \(\frac{t}{10}\)

∴ t = 50 yr

Section – C

Question 16.

Explain how Biot-Savart law enables one to express the Ampere’s circuital law in the integral form, viz. [2 + 1 = 3]

\(\oint B \cdot d l\) = μ_{0}I

where, I is the total current passing through the surface.

Or

Darw a labelled diagram of a moving coil galvanometer and explain its working. What is the function of radial magnetic field inside the coil? [2 + 1 = 3]

Answer:

Consider any arbitrary closed path perpendicular to the plane of paper around a long straight conductor XY carrying current from X to Y, lying in the plane of paper.

Let, the closed path be made of large number of small elements, where

AB = dl_{1}, BC = dl_{2}, CD = dl_{3}

Let dθ_{1}, dθ_{2}, dθ_{3}, be the angles subtended by the various elements at point O through which conductor is passing.

Then

dθ_{1} + dθ_{2} + dθ_{3} + …….. = 2π

Suppose these small elements AB, BC, CD, are small circular arcs of radii r_{1}, r_{2}, r_{3}, ….respectively.

Then, dθ_{1} = \(\frac{d I_{1}}{r_{1}}\), dθ_{2} = \(\frac{d I_{2}}{r_{2}}\),

dθ_{3} = \(\frac{d I_{3}}{r_{3}}\)

If B_{1}, B_{2}, B_{3} are the magnetic field induction at a point along the small elements dl_{1}, dl_{2}, dl_{3}, then from Biot-Savart’s law we know that for the conductor of infinite length, magnetic field is given by

B_{1} = \(\frac{\mu_{0}}{4 \pi} \frac{2 I}{r_{1}}\)

B_{2} = \(\frac{\mu_{0}}{4 \pi} \frac{2 I}{r_{2}}\)

B_{3} = \(\frac{\mu_{0}}{4 \pi} \frac{2 I}{r_{3}}\)

In case of each elements, the magnetic field induction B and current element vector dl are in the same direction. Line integral of B around closed path is

⇒ \(\) = μ_{0}I, which is an expression of Ampere’s circuital law.

Question 17.

With the help of a suitable ray diagram, derive a relation between the object distance (u), image distance (v) and radius of curvature R for the convex spherical surface when a ray of light travels from a rarer to denser mediuht. [1 + 2 = 3]

Or

Draw a schematic ray diagram of reflecting telescope showing how rays coming from a distant object are received at the eyepiece. Deduce expression for magnifying power. [1 + 2 = 3]

Answer:

According to Snell’s law,

_{1}n_{2} = \(\frac{\sin i}{\sin r}\)

Where _{1}n_{2} = the refractive index of medium (2) with respect to medium (1).

or n = \(\frac{\sin i}{\sin r}\)

(on taking n in place of _{1}n_{2} temporarily) If the angles i and r are very small, then

sin i ≈ i and sin r = r ≈ r

∴ n = \(\frac{i}{r}\)

or i = nr ……………(1)

∴ In triangle, the exterior angle is equal to sum of the opposite two interior angles.

∴ From ΔMOC, i = α + γ ………….. (2)

And from ΔMIC,

r = β + γ ………………. (3)

Substituting the value of i and r in equation (1), then we get,

(α + γ) = n(β + γ) ………………. (4)

If point M is not far from the principal axis, then

(i) Points P and P’ are very close to each other and considered to be same point.

(ii) Angles α, β and γ will be small.

Therefore, α = tanα = \(\frac{M P}{O P}=\frac{M P}{-u}\)

β ≈ tanβ = \(\frac{M P}{I P}=\frac{M P}{-v}\)

and γ ≈ tanγ = \(\frac{\mathrm{MP}}{C P}=\frac{M P}{R}\)

Now, substituting the values of α, β and γ in equation (4) then, we get

Now again writing _{1}n_{2} in place of n, we get

\(\frac{n_{2}}{v}-\frac{1}{u}=\frac{1 n_{2}-1}{R}\) ……………… (5)

This formula is known as

‘Refraction Formula’ of convex surface.

Question 18.

(a) Quarks inside protons and neutrons are thought to carry fractional charges (+\(\frac{2}{3}\)e,- \(\frac{1}{3}\)e) Why do they not show up in millikan’s oil drop experiment?

(b) Why do we need the oil drops of Millikan’s experiment to be of such microscopic sizes? Why cannot we experiment with much bigger drops?

(c) Stoke’s formula for viscons drag is not really valid for oil drops of extremely minute sizes. Why not? [1 + 1 + 1 = 3]

Or

(i) Why photoelectric effect cannot be explained on the basis of wave nature of light? Give reasons.

(ii) Write the basic features of photon picture of electromagnetic radiation on which Einstein’s photoelectric equation is based. [1 + 1 = 3]

Answer:

(a) Quarks are thought to be confined within a proton or neutron by forces which grow stronger if one tries to pull them apart. It, therefore, seems that though fractional charges may exist in nature, observable changes are still integral multiples of electronic charge ‘e’.

(b) Electric fields needed in the experiment will be impractically high.

(c) Stokes’ formula is valid for motion through a homogeneous continuous medium. The size of the drop should be much larger than the intermolecular separation in the medium for this assumption to be valid; athomise the drop ‘sees’ inhomogeneities in the medium, i.e. there is concentrated mass density in the molecules, and voids in between the molecules.

Section – D

Question 19.

(i) Use Gauss’s theorem to find the electric field due to a uniformly } charged infinitely large plane thin sheet with surface charge density σ.

(ii) An infinitely large thin plane sheet has a uniform surface charge density σ. Obtain the expression for the amount of work done in bringing a point charge q from infinity to a point, distant r, in front of the charged plane sheet. [1 + 2 + 1 = 4]

Or

Obtain the formula for the electric field due to a long thin wire of uniform linear charge density X without using Gauss’s law. [2 + 2 = 4]

Answer:

Direction of electric field due to an infinite charged sheet:

Formula for Electric Field:

To determine the electric field at a point P, we draw a Gaussian surface. According to figure we draw a cylindrical Gaussian surface of area of cross-section S such that its one place surface S_{1} passes through P, and the plane. Surface S_{2} is on the other side of the charge sheet.

The enclosed charge, Σq = σS

where σ is the surface charge density

∴ From Gauss’s law,

Φ_{E} = \(\frac{\sigma S}{\varepsilon_{0}}\) …………… (1)

From the definition of flux,

= ES + ES

or Φ_{E} = 2ES ………………. (2)

Comparing equation (1) and (2)

2ES = \(\frac{\sigma S}{\varepsilon_{0}}\)

∴ E = \(\frac{\sigma}{2 \varepsilon_{0}}\) ………………. (3)

In vector form,

\(\vec{E}=\frac{\sigma}{2 \varepsilon_{0}} \hat{n}\) …………….. (4)

(ii) Surface charge density of the uniform plane sheet which is infinitely large = + α. The electric potential (V) due to infinite sheet of uniform charge density + σ

V = \(\frac{\sigma r}{2 \varepsilon_{0}}\)

The amount of work done in bringing a point charge q from infinite to point, at distance r in front of the charged plane sheet.

W = q × V = q.\(\frac{-\sigma r}{2 \varepsilon_{0}}\) = – \(\frac{\sigma r \cdot q}{2 \varepsilon_{0}}\)J

Question 20.

(i) Draw I-V characteristics of a Zener diode.

(ii) Explain with the help of a circuit diagram, the use of a Zener diode as a voltage regular.

(iii) A photodiode is operated under reverse bias although in the forward bias,

the current is known to be more than the current in the reverse bias. Explain, giving reason. [1 + 2 + 1 = 4]

Or

(i) How is forward biasing different from reverse biasing in a p-n junction diode?

(ii) Draw the circuit diagram showing how a p-n junction diode is: (a) forward biased (b) reverse biased. [2 + 2 = 4]

Answer:

Figure shows a power supply whose output voltage is assumed to be V_{i} and a resistance R_{s} and a zener diode are connected at its output terminals. A load resistor R_{L} is parallel to the zener diode. The zener diode is so selected whose breakdown voltage V_{Z} is equal to the constant value of that dc voltage which is to be obtained at load resistor R_{L}. Zener diode is so placed in the circuit that it is in reverse biasing state. If the value of V_{i} is more than V_{Z} then the diode comes in breakdown region and the potential V_{Z} remains at its ends. Since R_{L} is parallel to the diode, hence on it also there would be V_{Z} potential. If the value of R_{L} is changed then also it will have V_{Z} potential. In this way the output voltage at both the ends of a zener diode is stable. The resistance Rs is so selected that the diode is not damaged by the thermal energy due to the increase in the zener current,

(iii) In n -type semiconductor,

n_{e} > n_{h} …………….. (i)

On incidence of light of suitable frequency, there is equal rise in number of electrons and holes [i.e., Δn (say)]

⇒ \(\frac{1}{n_{e}}<\frac{1}{n_{h}}\) or \(\frac{\Delta n}{n_{e}}<\frac{\Delta n}{n_{h}}\)

where, Δn = change in electron or hole charge carrier, Thus, fractional change in minority charge carrier (hole) is much higher than fraction change in majority charge carrier (electron). Also, minority charge carrier contribute in drift current in reverse bias. Thus, with incidence of light, fractional change in minority charge carrier is significant. Therefore, photodiode shoud be connected in reverse bias for measuring light intensity.

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