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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

1851.

A particle 'P' is moving in a circle of radius 'r' with uniform speed v. AB is the diameter of circle and 'C' is the centre. The angular velocity of Pabout A and C are in the ratio.

Answer»

`1:1`
`1:2`
`2:1`
`4:1`

ANSWER :B
Discussion
1852.

What will be the change in electric potential energy of a positive test charge q_(0) when it is displaced in a uniform electric field E=barE_(0)j from y_(i)= a to y_(r ) = 2a along the y-axis?

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ANSWER :`-q_(0)E_(0)a`
Discussion
1853.

What are permanent magnets ? What is an efficient way of preparing a permanent magnet ? Write two characteristic properties of materials which are required to select them for permanent magnets.

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Solution :SUBSTANCES which at room temperature retain their ferromagnetic PROPERTY for a long period of time are called "permanent magenet". An efficient way to make a permanent magnet is to place a ferromagnetic rod in a SOLENOID and pass a current. The MAGNETIC field of the solenoid magnetises the rod. Materials selected for PREPARING permanent magnet should have
(a) high retentivity so that the magnet is a strong magnet, and
(b) high coercivity so that the magnetisation is not erased by stray magnetic fields, temperature fluctuations or minor mechanical damage.
(c) high permeability so that material becomes a strong magnet. Steel and alloys like Alnico, Cobalt steel and Ticonal are suitable materials for preparing permanent magnets.
Discussion
1854.

An object starting from rest has two forces acting on it: one performing 40 J of work and the other (friction) performing -20J. What is the final kinetic energy of this object?

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Solution :The total work DONE is (40J)+(-20J)=20J. So, by the work-energy THEOREM,
`W_("total")=DeltaK`, we have 20J=`DeltaK`. Since `DeltaK=K_(f)-K_(f)-K_(i)`, we FIND that `K_(f)=K_(i)+DeltaK=0J+20J=20J`.
Discussion
1855.

A battery of emf 10 V and internal resistance 3Omega is connected to a resistor. The current in the circuit is 0.5A. The terminal voltage of the battery when the circuit is closed is

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10 V
ZERO
8.5 V
1.5 V

SOLUTION :`V=xi-Ir=10-(0.5xx3)=8.5V`
Discussion
1856.

The magnetic induction at the centre of a current carrying circular of coil radius r, is

Answer»

DIRECTLY PROPORTIONAL to `R`
Inversely proportional `r`
Directly proportional to `r`
Inversely proportional to `r`

ANSWER :B
Discussion
1857.

A girl is swinging on a swing in the standing position. How will the period of swing be affected if she sits down ?

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It will become shorter
It will become longer
It will not CHANGE
It will become INFINITE.

ANSWER :B
Discussion
1858.

A capacitor stores a charge of amount 0.05 C when connected across a battery of 100 V for a sufficient time. How much energy will be relaeased by this capacitor when it is discharged?

Answer»

Solution :Given, charge STORED in the capacitor,`Q=0.05C`
POTENTAIL difference across the capacitor, `V=100V`
ENERGY stored in the capacitor will be:
`U=(1)/(2) QV`
`=(1)/(2)xx0.05xx100`
`U=2.5J`
The same energy will be RELEASED when the capacitor is discharged.
Discussion
1859.

Infrared radiation is detected by

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spectrometer
pyrometer
nanometer
photometer

Answer :B
Discussion
1860.

The average American film is a …………… model.

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GOOD
BAD
AVERAGE
NONE OF THE ABOVE

Answer :B
Discussion
1861.

A black body is at temeprature 300 K. It emits energy at a rate which is proportional to :

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300
`(300)^(3)`
`(300)^(2)`
`(300)^(4)`

Solution :By Stefan.s LAW `EpropT^(4)`
`rArrE_(2)PROP(300)^(4)`
Correct CHOICE is (d).
Discussion
1862.

In an oscillating LC circuit, the maximum charge on the capacitor is Q. The charge on the capacitor when the energy is stored equally between the electric and magnetic fields is

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`(Q)/(2)`
`(Q)/(SQRT(3))`
`(Q)/(sqrt(2))`
Q

Answer :B
Discussion
1863.

According to Einstein's equation of photo-electric effect which of the following graph shows realtion between kinetic energy of electron emitted and frequency of incident radiation?

Answer»




Solution :`KE=HV-PHI`
`therefore` Graph of `KE to v` is a straight LINE .Its SLOPE is equal to h and K.E `gt` 0, whereas `hv gt phi`
Discussion
1864.

For the eletrostatic charge system as shown in (Fig. 3.121), find . a. the net force on electric dipole, and b. electrostatic energy of the system.

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Solution :a. Force exerted by the upper charge on dipole
`F_1 = (1)/(2 pi epsilon_0) (p q)/(a^3)` (down)
force extered by the left charge on dipole
`F_2 = (1)/(4 pi epsilon_0) (p q)/(a^3)` (up)
Force extered by the right charge on dipole
`F_3 = (1)/(4 pi epsilon_0) (p q)/(a^3)` (up)
Net force on the dipole due to all charges
`F = F_1 + F_2 + F_3 = 0`
Hence, net force on the dipole is zero. The TOTAL electric potential ENERGY consists of interaction of all the three charges among themselves and interaction of these three charges with dipole. So,
`U = 2 ((1)/(4 pi epsilon_0)(q^2)/(sqrt(2 a))) + (1)/(4 pi epsilon_0)(q^2)/(2 a)`
`- vec P . vec E _(up) -vec P . vec E _("left") - vec P . vec E _("right")`
`vec P . vec E_("left") = vec P . vec E_("right") = 0`
(Because electric fields PRODUCED by left and right charges are perpendicular to P.)
`- vec P . vec E_(up) = - P. ((1)/(4 pi epsilon_0) q/(a^2)) cos pi = (1)/(4 pi epsilon_0) (q p)/(a^2)`
B Putting the values, we get
`U = (1)/(4 pi epsilon_0) (q^2)/(2a) [ 2 sqrt(2) + 1] + (1)/(4 pi epsilon_0) (p q)/(a^2)`.
Discussion
1865.

Two poles-one 4 m high and the other is 4.5 m high are situated at distance 40 m and 50 m respectively from an eye. Which pole will appear taller

Answer»

SOLUTION :4 m POLE
Discussion
1866.

A parallel plate capacitor has plate area A and separation d. It is charged to a potential difference V. The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is

Answer»

`(3 epsi_0AV_(0)^(2))/( d) `
`(epsi_0 AV_(0)^(2))/( 2 d)`
`(epsi_0 AV_(0)^(2) )/(3d)`
`(epsi_0 AV_(0)^(2) )/( d)`

ANSWER :D
Discussion
1867.

By giving explanation of electric current, define current and its SI unit.

Answer»

Solution :`rArr` Due to motion of charges in conductor electric current is formed.
In a cross-section of a conductor let charge Q + flows in time t in forward direction (in direction of electric field and q - charge flows in backward direction in the similar time t.
`rArr ` Net charge flowing in forward direction in given cross-section in time t,
q = q + - q -
`rArr` Electric current (Definition) : Net flow of charge through cross-section of a conduction in un time is called electric current (I).
`rArr `When flow of charge changes with time, the electric current is defined as follows :
`rArr` ln time interval between t and t`Delta`tlet nc charge flowing through cross-section conductor be `Delta`Q ,then average current,
`rArr "" lt I gt = (Delta Q)`
by taking `underset(Delta t rarr 0)(" lim ")` we get instantaneous current ,
`I = underset(Delta t rarr 0 )( " lim ") (Delta Q)/(Delta t) = (d Q)/(dt)`
`therefore I = (dQ)/(dt)`
`rArr` Conventional current is taken in direction o motion of positive charge. In conductor positivt charge do not flow. Current is produced due to motion of electron. Hence, direction of current taken in opposite direction of motion of electron
`rArr` From I = `(q)/(t)` unit of electric current will be `("Coulomb")/("Second")`.
`rArr` In SI system unit of current is Ampere.
`rArr ` Definition of unit of electric current :
If in a cross-section of a conductor charge flowing in 1 second is 1 Coulomb `(6.25 xx 10^(18) "ELELCTRON")` then current flowing through is called 1 Ampere.
value of electric current in devices used in day to day life is of ORDER of ampere. Current flowing in LIGHTENING is oder of ten thousand ampere, current flowing in nerves of our body is order of `MU` A (microapmere).
Discussion
1868.

षट्कोणीय द्विविमीयनिबिड़ संकुल में C.N. कितना है -

Answer»

2
4
6
8

Answer :C
Discussion
1869.

A stone is dropped into a lake from a tower 500 metre high. The sound of the splash will be heard by the man approximately after :

Answer»

10 seconds
14 seconds
21 seconds
11.5 seconds.

Solution :t = `sqrt((2H)/(g) ) + (H)/(v)`
where v = 330 m/s is speed of sound
` = sqrt((2 xx 500)/(10) ) + (500)/(330) = 10 + 1.5 = 11. 5 ` seconds.
correct choice is (d) .
Discussion
1870.

A charged particle having mass m and charge q is accelerated by a potential difference V, it flies through a uniform transverse magnetic field B. The field occupies a region of space d. Fnd the time interval for which it remains inside the magnetic field.

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SOLUTION :`t=m(alpha)/(QB),"where "alpha=sin^(-1)((dBsqrt(q))/(SQRT(2mV)))` if `dltR,T//2` if `dgtR`
Discussion
1871.

A box is put on a scale which is adjusted to read zero, when the box is empty. A stream of pebbles is then poured into the box from a height h above its bottom at a rate of n pebbles. Each pebble has a mass m. If the pebbles collide with the box such that they immediately come to rest after collision, find the scale reading at time after the pebbles begin to fill the box.

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ANSWER :`{ SQRT((2gh) + GT))}`
Discussion
1872.

A bullet of mass 10 g strikes a ballistic pendulum of mass 2.0 kg. The center of mass of the pendulum rises a vertical distance of 12 cm. Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed.

Answer»


ANSWER :`3.1xx10^2` m/s
Discussion
1873.

An electron revolves in a circular loop with a frequency of 6 xx 10^15 cps. The current in the loop is

Answer»

0.96 mA
`0.96 mu A`
`28.8 mA`
`0.96 A `

SOLUTION :Equivalent current `I = EV = (1.6 XX 10^(-19)) xx (6 xx 10^15) = 9.6 xx 10^(-4) A = 0.96 mA`
Discussion
1874.

Out of following set offorces the resultant which can not be zero

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10, 10, 10
10, 10, 20
10, 20, 30
10, 20, 40,

ANSWER :D
Discussion
1875.

A block sliding down a rough 45° inclined plane has half the velocity it would have had, the inclined plane been smooth. The coefficient of sliding friction between block and the inclined plane is

Answer»

`1/3`
`3/4`
`1/2sqrt2`
`8/9`

ANSWER :D
Discussion
1876.

A beam of light of wavelength 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. What is the distance between the first dark fringe on either side of the central bright fringe?

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Solution :For FIRST MINIMUM (n = 1) on EITHER SIDE of central maximum.
`sin theta = (lambda)/(a)`
Where a is width of the slit
Since `theta` is very small. `(sin theta approx theta)`
`theta = (lambda)/(a)`
`sin theta approx theta = (x)/(2D)`
Where, x is distance of first dark fringe from central maximum.
D is distance between slit and screen.
From equation (1) and (2)
`(x)/(2D) = (lambda)/(a)`
`x=(2Dlamda)/(a)=(2xx2xx600xx10^(-9))/(1xx10^(-3))`
`=2400 xx 10^(-6) = 2.4 xx 10^(-3) m`
x = 2.4 mm
Discussion
1877.

A220 V A.C. supply is connectedbetweeen points A and B (See figure). What will be the potential difference V acorss the capacitor ?

Answer»

200V
110 V
0 V
`220 sqrt(2)` V

Solution :`220 sqrt(2) V`
In the given figure, applying Kirchhoff.s second law in the closed loop containing A.C. source, IDEAL diode and capacitor, we can write at time t,
`V=V_(D)+V_(C )`
`V_(m)SIN(omega t)=IR_(D)+V_(C )"" …(1)`
Resistance of ideal diode in forward BIASED condition is `R_(D)=0 ` and so,
`V_(C )=V_(m)sin (omega t)`
`therefore (V_(C ))_("max")=V_(m)`
`=V_("rms")xx sqrt(2)(because V_("rms")=(V_(m))/(sqrt(2)))`
`=220sqrt(2)` volt
Resistance of ideal diode in reverse biased condition is `R_(D)=oo` and so taking I = 0, from equation (1),
`V_(m) sin (omega t)=V_(C )`
`therefore (V_(C ))_("max")=V_(m)`
`=220sqrt(2)` volt
Discussion
1878.

In the above problem. The plane of the coil is initially kept parallel to B. The coil is rotated by an angle theta about the diameter perpendicular to B and charge of amount Q flows through it. Choose the correct alternatives.

Answer»

(a) `THETA = 90^(@), Q = (Ban//R)`
(b) `theta = 180^(@), Q = (Ban//R)`
( c) `theta = 180^(@), Q = 0`
( d) `theta = 360^(@), Q = 0`

SOLUTION :(a,b,d):
Charge flowing in thecircuit `= (Deltaphi)/(R )`
where, `Deltaphi` = CHANGE in flux
`= phi_(final) - phi_(initial)`
and`R` = RESISTANCE in the circuit
Discussion
1879.

The rate of outflow of liquid through an orifice does not depends upon

Answer»

RADIUS of orifice
Height of LIQUID column
Acceleration DUE to gravity
Density of liquid

Answer :C
Discussion
1880.

A dielectric in the form of a sphere is introduced into a homogeneous electric field. A, B and Care three points as shown in the below figure

Answer»

intensity at A increases while that at B and C DECREASES 
intensity at A and B decreases, where as intensity at C increases 
intensity at A and C increases and THA at B decreases 
intensity at A, B and C decreases 

ANSWER :C
Discussion
1881.

Unpolarised light is incident on a plane glass surface. What should be the angle of incidence so that the reflected and refracted rays are perpendicular to each other? (Given n=1.5)

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SOLUTION :for i+r to be equal to `(pi)/2`, we should have `tani_(B)=mu=1.5`. This GIVES `i_(B)=57^(@)`. This is theBrewster.s ANGLE for AIR to glass interface.
Discussion
1882.

Modulation index in FM signal

Answer»

VARIES inversely as the frequency DEVIATION
varies directly as the modulating frequency
varies directly as the frequency deviation
varies inversely as the SQUARE of the modulating frequency.

Answer :D
Discussion
1883.

A long straight wire carries a current of 4A. A proton travels with a velocity of 4 xx 10^4 ms-1 parallel to the wire 0.2m from it and in a direction opposite to the current. What is the force which the magnetic field due to current exerts on the moving proton?

Answer»

SOLUTION :`2.56 XX 10^(-20)`N
Discussion
1884.

The visible portion of a lightning strike is preceded by an invisible stage in which a column of electrons extends downward from a cloud to the ground Assume the linear chargedensity along the column is 1.00xx10^(-3)C//m Treat the column of charge as if it were straight and infinitely long. At what distance (in m) from the column of electrons does the electric field havea magnitude of 3.00xx10^(6) V//m the dielectric strength for air? This is an estimate of the radius of a visible lightning bolt. (Round off to nearest integer) (value of in_(0)is 8.85xx10^(-12) C^(2)//Nm^(2))

Answer»


SOLUTION :`lambda/(2pi in_(0)R)=E_(break)`
`r=lambda/(2piin_(0)E_(break))`
`=10^(-3)/(2xx3.14xx8.85xx10^(-12)xx3xx10^(6))`
`=1/(2xx3.14xx8.85)xx10^(3)`
`=5.99m APPROX 6m`
Discussion
1885.

Find the acceleration of the pulley after 5s, if the system is released from rest.

Answer»

SOLUTION :`12.5rad//s`
Discussion
1886.

Let is consider the following diagram in which a block of mass M is being supported by a uniform rope of mass 1 kg and length 10 metre, a pulse is created at the bottom of the rope and it reaches th etop. In column I the value of M in kg is given and in column II time (in sec) after which the pulse reaches the top is given. Match them:

Answer»


ANSWER :A-s; B-r; C-p; D-q
Discussion
1887.

Which discovery Faraday made public ? Discuss importance of electromagnetic induction.

Answer»

Solution :Faraday made public his discovery that relative motion between a bar MAGNET and a wire loop produced a small current in the latter.
The PHENOMENON of electromagnetic induction is not merely of theoretical or academic interest but also of practical utility. Imagine a world where there is no electricity - no electric lights, no trains, no telephones and no personal computers. Electromagnetic induction is the base of INVENTION of electricity.
The pioneering experiments of Faraday and Henry have led directly to the DEVELOPMENT of MODERN day generators and transformers.
Today.s civilization owes its progress to a great extent to the discovery of electromagnetic induction.
Discussion
1888.

(A) Its is not possible that all the lines avail-able in the emission spectrum will also be available in the absorption spectrum (R) The spectrum of hydrgen atom is only absorption spectrum

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Both 'A' and 'R' are TRUE and 'R' is the correct EXPLANATION of 'A'.
Both 'A' and 'R' are true and 'R' is not the correct explanation of 'A'
'A' is true and 'R' is false
'A' is false and 'R' is true

ANSWER :C
Discussion
1889.

In an n-p-n transistor circuit , the collector current is 10 mA . If 90% of the electrons emitted reach the collector , the emitter current I_e and base current I_b are given be :

Answer»

`I_e = 9MA , I_b=-1MA`
`I_e = -1mA , I_b=9mA`
`I_e = 1mA , I_b=11mA`
`I_e = 11mA , I_b=1mA`

SOLUTION :`i_c=10 mA= 0.9 I_e`
`i_e=10/(0.9)=11 mA`
Now , `i_e=i_c+i_b`
`:. i_b=I_e-i_c=11-10=1mA`
Discussion
1890.

The battery is charged from full wave rectifier fed by a sinuoidal voltage (see figure). Idea diodes, ammeter and voltmeter show the time average value. At idle with only key K_(1) closed, coltmeter shows 12 V, and current is then absent, ie, reading of ameter is 0. if only the key K_(2) is closed, the voltmeter shows battery voltage at 12.3 V. during charging when the K_(1) and K_(2) are closed, the voltmeter shows 12.8 V and ammeter shows 5 A. Find the internal resistance of battery.

Answer»


ANSWER :`0.1Omega`
Discussion
1891.

A point source of monochromatic light is positioned in front of a zone plate at a distance a = 1.5 m from it. The image of the source is formed at a distance b = 1.0 m from the plate. Find the focal length of the zone plate.

Answer»

Solution :In a zone plate an undarkened circular disc is FOLLOWED by a number of altermetely undarkened and darkened rings. For the proper case, correspond to `1^(st), 2^(nd), 3^(rd)……` FRESNEL zones.
LET `r_(1) =` radius of the central undarkened circle. Then for this to be first Fresnel zone in the present case, we must have
`SL + LI - SL = lambda//2`
Thus if `r_(1)` is the radius of the perphery of the first zone
`sqrt(a^(2) + r_(1)^(2)) + sqrt(b^(2) + r_(1)^(2)) -(a + b) = (lambda)/(2)`
or `(r_(1)^(2))/(2) ((1)/(a) + (1)/(b)) = (lambda)/(2)` or `(1)/(a) + (1)/(b) = (1)/(r_(1)^(2)//lambda)`
It is clear that the plate is acting as a lens of focal length
`f_(1) = (r_(1)^(2))/(lambda) = (ab)/(a+b) = 6`metre.
This is the principle focal length.
Other maxima are OBTAINED when
`SL + LI - SI = 3(lambda)/(2), 5(lambda)/(2),..........`
These focal lengths are ALSO `(r_(1)^(2))/(3lambda), (r_(1)^(2))/(5lambda),....`
Discussion
1892.

A photon with energy exceeding eta =2.0 times the rest energy of an electron experienced a head-on collision with a stationary free electron. Find the curvature radius of the trajectory of the compton electron in a magnetic field B = 0.12 T. The compton electron is assumed to mive at right angles to the direction of the field.

Answer»

SOLUTION :By head on collision we understand that the electron moves on in the direction of the incident photon after the collision and the photon is scaltered backwards. Then, let us write.
`cancel h omega = ETA mc^(2)`
`cancel h omega' = sigma mc^(2)`
`(E,p) = (epsilonc^(2), MU mc)` of the electron.
The by energy momentum consrvation (cancelling factors of `mc^(2)` and `mc`)
`1+eta = sigma + epsilon`
`eta = mu - sigma`
`epsilon^(2) = 1+mu^(2)`
So eliminating `sigma& epsilon` `1+eta =- eta+ mu + sqrt(mu^(2) +)`
or `(1+2eta - mu) = sqrt(mu^(2) +1)`
Squaring `(1+2 eta)^(2) - 2mu (1+2eta) = 1`
`4eta + 4eta^(2) = 2mu (1+2eta)`
or `mu = (2eta(1+eta))/(1+2eta)`
Thus the momentum of the Compton electron is
`p = mu c = (2eta(1+eta)mc)/(1+2eta)`.
Now in a megnetic FIELD `p = Be rho`
Thus `rho = 2 eta (1+eta)//(1+2eta) (mc)/(Be)`.
Substituting the values `rho = 3.412cm`.
Discussion
1893.

In Fig. 25-25 a potential difference V=75.0 V is applied across a capacitor arrangement with capacitances C_1=10.0mu F, C_2=5.00 mu F, and C_3=15.0 mu F. What are (a) charge q_3 (b) potential difference V_3 and ( c) stored energy U_3 for capacitor 3, (d) q_1 (e) V_1 and (f) U_1 for capacitor 1, and (g) q_2 (h) V_2 and (i) U_2 for capacitor 2?

Answer»

Solution :`a) 5.625 times 10^-4 C approx 563C B) 37.5V c) 10.5 MJ d) 37.5 times 10^-4 C e) 37.5V f) 7.03 mJ g) 1.88 times 10^-4 C h) 37.5 V i) 3.52 mJ`
Discussion
1894.

The filament of a bulb has an area 5xx10^(-5)m^(2) and is at 2000 K. If its relative emittance is 0.85 and sigma=5.7xx10^(-8) MKS units, what the energy radiated in one minuts :

Answer»

2336 J
`232.6` J
`23.26` J
None of the above

Solution :`E=eAsigmaT^(4)t=0.85xx5xx10^(-5)XX5*7XX10^(-8)(2000)^(4)xx60`.
`=2360J`.
THUS correct choice is (a).
Discussion
1895.

Relativistic corrections become necessary when the expression for kinetic energy (1)/(2) mv^(2) becomes comparable to mc^(2). At what de Broglie wavelength will relativistic corrections become important for an electron?

Answer»

10nm
`10^(-1)NM`
`10^(-4)nm`
`10^(-6)nm`

Answer :C::D
Discussion
1896.

A small flat search coil of area 2.0 cm^(2) with 25 close turns placed between the poles of a strong magnet normally to the magnetic field is suddenly sntached out of the field. The total charge flown in the coil as measured by a ballistic galvanometer connected to the coil, is 7.5mC. The resistance of the coil and the galvanometer is 0.50Omega. Find teh magntidue of the field of the magnet.

Answer»

SOLUTION :`0.75 Wb//m^(2)`
Discussion
1897.

The height of a transmitting antenna is 200m. Radius of earth is 6.4*10^6m. Find the range upto which the above antenna gives transmission signal.

Answer»

SOLUTION :`d_m= sqrtRh
d_m = sqrt2x6.410^6x200 m`
Discussion
1898.

What is the effect on the interference fringes in a Young.s double-slit experimental due to each of the following operations : The monochromatic source is replaced by another (monochromatic) source of shorter wavelength.

Answer»

Solution :The SEPARATION of the fringes (and also angular separation) decreases. SEE, however, the CONDITION mentioned in (d) below.
Discussion
1899.

What is the effect on the interference fringes in a Young.s double-slit experimental due to each of the following operations : The separation between the two slits is increased.

Answer»

SOLUTION :The separation of the fringes (and also ANGULAR separation) DECREASES. See, however, the condition MENTIONED in (d) below.
Discussion
1900.

What is the effect on the interference fringes in a Young.s double-slit experimental due to each of the following operations : The source slit is moved closer to the double-slit plane.

Answer»

SOLUTION :Let s be the size of the source and S its distance from the plane of the two SLITS. For interference fringes to be seen, the condition `s"/"S lt lambda"/"d` should be SATISFIED, otherwise, interference patterns produced by different parts of the source overlap and no fringes are seen. Thus, as S DECREASES (i.e., the source slit is brought closer), the interference pattern gets LESS and less sharp, and when the source is brought too close for this condition to be valid, the fringes disappear. Till this happens, the fringe separation remains fixed.
Discussion