MCQ 2011 Mark
Two Nicols are oriented with their principal planes making an angle of $60^{\circ}$. The percentage of incident unpolarized light which passes through the system is
- A
$50 \%$
- B
$100 \%$
- ✓
$12.5 \%$
- D
$37.5 \%$
AnswerCorrect option: C. $12.5 \%$
Intensity of polarized light from first polarizer
$=\frac{100}{2}=50$$I=50 \cos ^2 60^{\circ}=\frac{50}{4}=12.5$
View full question & answer→MCQ 2021 Mark
When a plane polarised light is passed through an analyser and analyser is rotated through $90^{\circ}$, the intensity of the emerging light
- A
Varies between a maximum and minimum
- B
- C
- ✓
Varies between a maximum and zero
AnswerCorrect option: D. Varies between a maximum and zero
View full question & answer→MCQ 2031 Mark
Polarising angle for water is $53^{\circ} 4^{\prime}$. If light is incident at this angle on the surface of water and reflected, the angle of refraction is
AnswerCorrect option: C. $36^{\circ} 56^{\prime}$
$\theta_P+r=90^{\circ}$ or $r=90-\theta=90^{\circ}-53^{\circ} 4^{\prime}=36^{\circ} 56^{\prime}$.
View full question & answer→MCQ 2041 Mark
The angle of incidence at which reflected light is totally polarized for reflection from air to glass (refraction index $n$ ) is
AnswerCorrect option: D. $\tan ^{-1}(n)$
$\mu=\tan \theta \Rightarrow \theta=\tan n$
View full question & answer→MCQ 2051 Mark
If $c$ is the speed of electromagnetic waves in vacuum, its speed in a medium of dielectric constant $K$ and relative permeability $\mu_r$ is
- A
$v=\frac{1}{\sqrt{\mu_r K}}$
- B
$v=c \sqrt{\mu_r K}$
- ✓
$v=\frac{c}{\sqrt{\mu_r K}}$
- D
$v=\frac{K}{\sqrt{\mu_r C}}$
AnswerCorrect option: C. $v=\frac{c}{\sqrt{\mu_r K}}$
Speed of light of vacuum $c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$ and in another medium$v=\frac{1}{\sqrt{\mu \varepsilon}}$
$\therefore \frac{c}{v}=\sqrt{\frac{\mu \varepsilon}{\mu_0 \varepsilon_0}}=\sqrt{\mu_r K} \Rightarrow v=\frac{c}{\sqrt{\mu_r I}}$
View full question & answer→MCQ 2061 Mark
Which of the following is electromagnetic wave
- ✓
$X$-rays and light waves
- B
Cosmic rays and sound waves
- C
Beta rays and sound waves
- D
Alpha rays and sound waves
AnswerCorrect option: A. $X$-rays and light waves
View full question & answer→MCQ 2071 Mark
The pressure exerted by an electromagnetic wave of intensity $l$ (watts $/ m)$ on a nonreflecting surface is [ $c$ is the velocity of light]
- A
$I c$
- B
$I c^2$
- ✓
$I / c$
- D
$I / c^2$
AnswerCorrect option: C. $I / c$
View full question & answer→MCQ 2081 Mark
The wavelength $21 cm$ emitted by atomic hydrogen in interstellar space belongs to
View full question & answer→MCQ 2091 Mark
A TV tower has a height of $100\ m$. The average population density around the tower is $1000$ per $km$. The radius of the earth is $6.4 \times 10^6 m$. the population covered by the tower is
- A
$2 \times 10^6$
- B
$3 \times 10^6$
- ✓
$4 \times 10^6$
- D
$6 \times 10^6$
AnswerCorrect option: C. $4 \times 10^6$
$\text { Population covered }=2 \pi h R \times \text { Population density } $
$=2 \pi \times 100 \times 6.4 \times 10 \times \frac{1000}{\left(10^3\right)^2}=4 \times 10^6$
View full question & answer→MCQ 2101 Mark
Television signals broadcast from the moon can be received on the earth while the TV broadcast from Delhi cannot be received at places about $100 km$ distant from Delhi. This is because
- A
There is no atmosphere around the moon
- B
Of strong gravity effect on TV signals
- ✓
TV signals travel straight and cannot follow the curvature of the earth
- D
There is atmosphere around the earth
AnswerCorrect option: C. TV signals travel straight and cannot follow the curvature of the earth
View full question & answer→MCQ 2111 Mark
A radio receiver antenna that is $2 m$ long is oriented along the direction of the electromagnetic wave and receives a signal of intensity $5 \times 10^{-16} W / m ^2$. The maximum instantaneous potential difference across the two ends of the antenna is
- ✓
$1.23 \mu V$
- B
$1.23 mV$
- C
$1.23 V$
- D
$12.3 mV$
AnswerCorrect option: A. $1.23 \mu V$
$I=\frac{1}{2} \varepsilon_0 C E_0^2 $
$\Rightarrow E_0=\sqrt{\frac{2 I}{\varepsilon_0 C}}=\sqrt{\frac{2 \times 5 \times 10^{-16}}{8.85}}=0.61 \times 10^{-6} \frac{ V }{ m }$
Also $E_0=\frac{V_0}{d}$
$ \Rightarrow V_0=E_0 d=0.61 \times 10^{-6} \times 2=1.23 \mu V$
View full question & answer→MCQ 2121 Mark
An $\text{LC}$ resonant circuit contains a $400pF$ capacitor and a $100 \mu H$ inductor. It is set into oscillation coupled to an antenna. The wavelength of the radiated electromagnetic waves is
- A
$377\ mm$
- B
$377$ metre
- ✓
$377 \ cm$
- D
$3.77 \ cm$
AnswerCorrect option: C. $377 \ cm$
$377 \ cm$
View full question & answer→MCQ 2131 Mark
An electromagnetic wave going through vacuum is described by $E=E_0 \sin (k x-\omega t) ; \quad B=B_0 \sin (k x-\omega t)$. Which of the following equation is true
- ✓
$E_0 k=B_0 \omega$
- B
$E_0 \omega=B_0 k$
- C
$E_0 B_0=\omega k$
- D
AnswerCorrect option: A. $E_0 k=B_0 \omega$
$\frac{E_0}{B_0}=C$. also $k=\frac{2 \pi}{\lambda}$ and $\omega=2 \pi v$
These relation gives $E_0 K=B_0 \omega$
View full question & answer→MCQ 2141 Mark
An electromagnetic wave, going through vacuum is described by $E=E_0 \sin (k x-\omega t)$. Which of the following is independent of wavelength
- A
$k$
- B
$\omega$
- ✓
$k / \omega$
- D
$k \omega$
AnswerCorrect option: C. $k / \omega$
The angular wave number $k=\frac{2 \pi}{\lambda}$; where $\lambda$ is the wave length.
The angular frequency is $w=2 \pi v$.
The ratio $\frac{k}{\omega}=\frac{2 \pi / \lambda}{2 \pi v}=\frac{1}{v \pi}=\frac{1}{c}=$ constant
View full question & answer→MCQ 2151 Mark
A plane electromagnetic wave is incident on a material surface. If the wave delivers momentum $p$ and energy $E$, then
- A
$p=0, E=0$
- ✓
$p \neq 0, E \neq 0$
- C
$p \neq 0, E=0$
- D
$p=0, E \neq 0$
AnswerCorrect option: B. $p \neq 0, E \neq 0$
EM waves carry momentum and hence can exert pressure on surfaces.
They also transfer energy to the surface so $p \neq 0$ and $E \neq 0$.
View full question & answer→MCQ 2161 Mark
Which of the following rays has the maximum frequency
Answer$v_{\gamma-\text { rays }}>v_{U V \text {-rays }}>v_{\text {Blue light }}>v_{\text {Infrared rays }}$
View full question & answer→MCQ 2171 Mark
It is believed that the universe is expanding and hence the distant stars are receding from us. light from such a star will show
- ✓
Shift in frequency towards longer wavelengths
- B
Shift in frequency towards shorter wavelength
- C
No shift in frequency but a decrease in intensity
- D
A shift in frequency sometimes towards longer and sometimes towards shorter wavelengths
AnswerCorrect option: A. Shift in frequency towards longer wavelengths
View full question & answer→MCQ 2181 Mark
light from the constellation Virgo is observed to increase in wavelength by $0.4 \%$. With respect to Earth the constellation is
- ✓
(a) Moving away with velocity $1.2 \times 10 m / s$
- B
(b) Coming closer with velocity $1.2 \times 10 m / s$
- C
(c) Moving away with velocity $4 \times 10 \times m / s$
- D
(d) Coming closer with velocity $4 \times 10^{\circ} m / s$
AnswerCorrect option: A. (a) Moving away with velocity $1.2 \times 10 m / s$
(a)$\text { Using } \frac{\Delta \lambda}{\lambda}=\frac{v}{c} $
$\Rightarrow v=\frac{\Delta \lambda}{\lambda} c $
$\Rightarrow v=0.004 \times 3 \times 10^8=1.2 \times 10^6 m / sec$
View full question & answer→MCQ 2191 Mark
When the wavelength of light coming from a distant star is measured it is found shifted towards red. Then the conclusion is
- A
The star is approaching the observer
- ✓
The star recedes away from earth
- C
There is gravitational effect on the light
- D
The star remains stationary
AnswerCorrect option: B. The star recedes away from earth
View full question & answer→MCQ 2201 Mark
A rocket is going away from the earth at a speed $0.2 c$, where $c=$ speed of light. It emits a signal of frequency $4 \times 10^7 Hz$. What will be the frequency observed by an observer on the earth
- A
$4 \times 10^6 Hz$
- ✓
$3.2 \times 10^7 Hz$
- C
$3 \times 10^6 Hz$
- D
$5 \times 10^7 Hz$
AnswerCorrect option: B. $3.2 \times 10^7 Hz$
$v^{\prime}=v\left(1-\frac{v}{c}\right)=4 \times 10^7\left(1-\frac{0.2 c}{c}\right)$
$=3.2 \times 10^7 Hz$
View full question & answer→MCQ 2211 Mark
Due to Doppler's effect, the shift in wavelength observed is $0.1 \mathring A$ for a star producing wavelength $6000 \mathring A$. Velocity of recession of the star will be
- A
$2.5 km / s$
- B
$10 km / s$
- ✓
$5 km / s$
- D
$20 km / s$
AnswerCorrect option: C. $5 km / s$
$\frac{\Delta \lambda}{\lambda}=\frac{v}{c}$.
$\therefore v=\frac{\Delta \lambda}{\lambda} c=\frac{0.1}{6000} \times 3 \times 10^5 km / s =5 km / s$
View full question & answer→MCQ 2221 Mark
A star moves away from earth at speed $0.8 c$ while emitting light of frequency $6 \times 10^{14} Hz$. What frequency will be observed on the earth (in units of $\left.10^* Hz \right)(c=$ speed of light)
Answer$\text { Observed frequency } v^{\prime}=v\left(1-\frac{v}{c}\right) $
$\Rightarrow v^{\prime}=6 \times 10^{14}\left(1-\frac{0.8 c}{c}\right)=1.2 \times 10^{14} Hz$
View full question & answer→MCQ 2231 Mark
A star emitting radiation at a wavelength of $5000 \mathring A$ is approaching earth with a velocity of $1.5 \times 10^6 m / s$. The change in wavelength of the radiation as received on the earth, is
- ✓
$25 \mathring A$
- B
- C
$100 \mathring A$
- D
$2.5 \mathring A$
AnswerCorrect option: A. $25 \mathring A$
$\Delta \lambda=\lambda \cdot \frac{v}{c}=\frac{1.5 \times 10^6}{3 \times 10^8} \times 5000=25 \mathring A$
View full question & answer→MCQ 2241 Mark
If a source of light is moving away from a stationary observer, then the frequency of light wave appears to change because of
AnswerAccording to Doppler's effect, wherever there is a relative motion between source and observer, the frequency observed is different from that given out by source.
View full question & answer→MCQ 2251 Mark
A star is moving away from the earth with a velocity of $100 km / s$. If the velocity of light is $3 \times 10^8 m / s$ then the shift of its spectral line of wavelength $5700 \mathring A$ due to Doppler's effect will be
- A
$0.63 \mathring A$
- ✓
$1.90 \mathring A$
- C
$3.80 \mathring A$
- D
$5.70 \mathring A$
AnswerCorrect option: B. $1.90 \mathring A$
$\Delta \lambda=\lambda \frac{v}{c}=5700 \times \frac{100 \times 10^3}{3 \times 10^8}=1.90 \mathring A$
View full question & answer→MCQ 2261 Mark
In the context of Doppler effect in light, the term 'red shift' signifies
MCQ 2271 Mark
A spectral line $\lambda=5000 \mathring A$ in the light coming from a distant star is observed as a $5200 \mathring A$. What will be recession velocity of the star
- A
$1.15 \times 10^7 cm / sec$
- ✓
$1.15 \times 10^7 m / sec$
- C
$1.15 \times 10^7 km / sec$
- D
$1.15 km / sec$
AnswerCorrect option: B. $1.15 \times 10^7 m / sec$
$\Delta \lambda=5200-5000=200 \mathring A$
Now $\frac{\Delta \lambda}{\lambda^{\prime}}=\frac{v}{c}$
$ \Rightarrow v=\frac{c \Delta \lambda}{\lambda^{\prime}}=\frac{3 \times 10^8 \times 200}{5000}$$=1.2 \times 10^7 m / sec \approx 1.15 \times 10^7 m / sec$
View full question & answer→MCQ 2281 Mark
A rocket is moving away from the earth at a speed of $6 \times 10^7 m / s$. The rocket has blue light in it. What will be the wavelength of light recorded by an observer on the earth (wavelength of blue light $=4600\ \mathring A$ )
- A
$4600 \mathring A$
- ✓
$5520 \mathring A$
- C
$3680 \mathring A$
- D
$3920 \mathring A$
AnswerCorrect option: B. $5520 \mathring A$
$\frac{\Delta \lambda}{\lambda} =\frac{v}{c}=\frac{6 \times 10^7}{3 \times 10^8}=0.2 $
$\Delta \lambda =\lambda^{\prime}-\lambda=0.2 \lambda $
$\Rightarrow \lambda^{\prime}=1.2 \lambda=1.2 \times 4600=5520 \mathring A$
View full question & answer→MCQ 2291 Mark
Assuming that universe is expanding, if the spectrum of light coming from a star which is going away from earth is tested, then in the wavelength of light
- A
- ✓
The spectrum will move to infrared region
- C
The spectrum will seems to shift to ultraviolet side
- D
AnswerCorrect option: B. The spectrum will move to infrared region
Due to expansion of universe, the star will go away from the earth thereby increasing the observed wavelength. Therefore the spectrum will shift to the infrared region.
View full question & answer→MCQ 2301 Mark
A Young's double slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen is
MCQ 2311 Mark
A beam of plane polarized light falls normally on a polarizer of cross sectional area $3 \times 10^{-4} m ^2$. Flux of energy of incident ray in $10 W$. The polarizer rotates with an angular frequency of $31.4 rad / sec$. The energy of light passing through the polarizer per revolution will be
Answer(b) For maxima $\Delta=d \sin \theta=n \lambda$
$\Rightarrow 2 \lambda \sin \theta=n \lambda$
$ \Rightarrow \sin \theta=\frac{n}{2}$
since value of $\sin \theta$ can not be greater $l$.
$\therefore n=0,1,2$
Therefore only five maximas can be obtained on both side of the screen.
$\text { (a) } \frac{\Delta \lambda}{\lambda}=\frac{v}{c} $
$\Rightarrow \frac{(401.8-393.3)}{393.3}=\frac{v}{3 \times 10^8} $
$\Rightarrow v=6.48 \times 10 m / s =6480 km / sec .$
View full question & answer→MCQ 2321 Mark
A beam of natural light falls on a system of $6$ polaroids, which are arranged in succession such that each polaroid is turned through $30^{\circ}$ with respect to the preceding one. The percentage of incident intensity that passes through the system will be
- A
$100 \%$
- B
$50 \%$
- C
$30 \%$
- ✓
$12 \%$
AnswerCorrect option: D. $12 \%$
$12 \%$
View full question & answer→MCQ 2331 Mark
A plane electromagnetic wave of wave intensity $6 W / m$ strikes a small mirror area $40 cm$, held perpendicular to the approaching wave. The momentum transferred by the wave to the mirror each second will be
- A
$6.4 \times 10^{-7} kg - m / s ^2$
- B
$4.8 \times 10^{-8} kg - m / s ^2$
- C
$3.2 \times 10^{-9} kg - m / s ^2$
- ✓
$1.6 \times 10^{-10} kg - m / s ^2$
AnswerCorrect option: D. $1.6 \times 10^{-10} kg - m / s ^2$
Momentum transferred in one second
$p=\frac{2 U}{c}=\frac{2 S_{a v} A}{c}=\frac{2 \times 6 \times 40 \times 10^{-4}}{3 \times 10^8} $
$=1.6 \times 10^{-10} kg - m / s .$
View full question & answer→MCQ 2341 Mark
A wave is propagating in a medium of electric dielectric constant 2 and relative magnetic permeability 50 . The wave impedance of such a medium is
- A
$5 \Omega$
- B
$376.6 \Omega$
- ✓
$1883 \Omega$
- D
$3776 \Omega$
AnswerCorrect option: C. $1883 \Omega$
Wave impedance $Z=\sqrt{\frac{\mu_r}{\varepsilon_r}} \times \sqrt{\frac{\mu_0}{\varepsilon_0}}$$=\sqrt{\frac{50}{2}} \times 376.6=1883 \Omega$
View full question & answer→MCQ 2351 Mark
A point source of electromagnetic radiation has an average power output of $800\ W$. The maximum value of electric field at a distance $4.0 m$ from the source is
- A
$64.7 V / m$
- B
$57.8 V / m$
- C
$56.72 V / m$
- ✓
$54.77 Vm$
AnswerCorrect option: D. $54.77 Vm$
Intensity of EM wave is given by$I=\frac{P}{4 \pi R^2}=v_{a v} \cdot c=\frac{1}{2} \varepsilon_0 E_0^2 \times c$
$\Rightarrow E_0 =\sqrt{\frac{P}{2 \pi R^2 \varepsilon_0 c}} $
$=\sqrt{\frac{800}{2 \times 3.14 \times(4)^2 \times 8.85 \times 10^{-12} \times 3 \times 10^8}} $
$ =54.77 \frac{V}{m}$
View full question & answer→MCQ 2361 Mark
A lamp emits monochromatic green light uniformly in all directions. The lamp is $3 \%$ efficient in converting electrical power to electromagnetic waves and consumes $100 W$ of power. The amplitude of the electric field associated with the electromagnetic radiation at a distance of $10 m$ from the lamp will be
- ✓
$1.34 V / m$
- B
$2.68 V / m$
- C
$5.36 V / m$
- D
$9.37 V / m$
AnswerCorrect option: A. $1.34 V / m$
$S_{a v}= \frac{1}{2} \varepsilon_0 c E_0^2=\frac{P}{4 \pi R^2} $
$\Rightarrow E_0 =\sqrt{\frac{P}{2 \pi R^2 \varepsilon_0 C}} $
$ =\sqrt{\frac{3}{2\times 3.14 \times 100 \times 8.85 \times 10^{-12} \times 3 \times 10^8}}$
$ =1.34 V / m$
View full question & answer→MCQ 2371 Mark
A laser beam can be focussed on an area equal to the square of its wavelength $A He - Ne$ laser radiates energy at the rate of $1 mW$ and its wavelength is $632.8\ nm$. The intensity of focussed beam will be
- A
$1.5 \times 10^{13} W / m ^2$
- ✓
$2.5 \times 10^9 W / m ^2$
- C
$3.5 \times 10^{17} W / m ^2$
- D
AnswerCorrect option: B. $2.5 \times 10^9 W / m ^2$
Area through which the energy of beam passes $=\left(6.328 \times 10^{-7}\right)=4 \times 10^{-13} m ^2 $
$\therefore I=\frac{P}{A}=\frac{10^{-3}}{4 \times 10^{-13}}=2.5 \times 10^9 W / m ^2$
View full question & answer→MCQ 2381 Mark
In an electromagnetic wave, the amplitude of electric field is $1 V / m$. the frequency of wave is $5 \times 10^{14} Hz$. The wave is propagating along $z$-axis. The average energy density of electric field, in Joule/m, will be
- A
$1.1 \times 10^{-11}$
- ✓
$2.2 \times 10^{-12}$
- C
$3.3 \times 10^{-13}$
- D
$4.4 \times 10^{-14}$
AnswerCorrect option: B. $2.2 \times 10^{-12}$
Average energy density of electric field is given by
$u_e=\frac{1}{2} \varepsilon_0 E^2=\frac{1}{2} \varepsilon_0\left(\frac{E_0}{\sqrt{2}}\right)^2=\frac{1}{4} \varepsilon_0 E_0^2 $
$=\frac{1}{4} \times 8.85 \times 10^{-12}(1)^2=2.2 \times 10^{-12} J / m ^3 .$
View full question & answer→MCQ 2391 Mark
A long straight wire of resistance $R$, radius $a$ and length $l$ carries a constant current $l$. The Poynting vector for the wire will be
AnswerCorrect option: D. $\frac{I^2 R}{2 \pi a l}$
Electric field $E=\frac{V}{l}=\frac{i R}{l}$ ( $R=$ Resistance of wire $)$Magnetic field at the surface of wire $B=\frac{\mu_0 i}{2 \pi a}$ ( $a=$ radius of wire)Hence poynting vector, directed radially inward is given by$S=\frac{E B}{\mu_0}=\frac{i R}{\mu_0 l} \cdot \frac{\mu_0 i}{2 \pi a}=\frac{i^2 R}{2 \pi a l}$
View full question & answer→MCQ 2401 Mark
A parallel plate capacitor of plate separation $2 mm$ is connected in an electric circuit having source voltage $400 V$. if the plate area is $60 cm$, then the value of displacement current for $10^{-6} sec$ will be
- A
$1.062 amp$
- ✓
$1.062 \times 10^{-2} amp$
- C
$1.062 \times 10^{-3} amp$
- D
$1.062 \times 10^{-4} amp$
AnswerCorrect option: B. $1.062 \times 10^{-2} amp$
$I_D=\varepsilon_0 \frac{d \phi_E}{d t}=\varepsilon_0 \frac{E A}{t}=\varepsilon_0\left(\frac{V}{d}\right) \cdot \frac{A}{t} $
$=\frac{8.85 \times 10^{-12} \times 400 \times 60 \times 10^{-4}}{10^{-3} \times 10^{-6}}=1.602 \times 10^{-2} amp$
View full question & answer→MCQ 2411 Mark
The ratio of the intensity at the centre of a bright fringe to the intensity at a point one-quarter of the distance between two fringe from the centre is
Answer(d) Electric field $E=\frac{V}{l}=\frac{i R}{l}$ ( $R=$ Resistance of wire $)$Magnetic field at the surface of wire $B=\frac{\mu_0 i}{2 \pi a}$ ( $a=$ radius of wire)Hence poynting vector, directed radially inward is given by$S=\frac{E B}{\mu_0}=\frac{i R}{\mu_0 l} \cdot \frac{\mu_0 i}{2 \pi a}=\frac{i^2 R}{2 \pi a l}$
View full question & answer→MCQ 2421 Mark
Among the two interfering monochromatic sources $A$ and $B ; A$ is ahead of $B$ in phase by $66^{\circ}$. If the observation be taken from point $P$, such that $P B-P A=\lambda / 4$. Then the phase difference between the waves from $A$ and $B$ reaching $P$ is
- ✓
$156^{\circ}$
- B
$140^{\circ}$
- C
$136^{\circ}$
- D
$126^{\circ}$
AnswerCorrect option: A. $156^{\circ}$
Total phase difference$=$ Initial phase difference + Phase difference due to path$=66^{\circ}+\frac{360^{\circ}}{\lambda} \times \Delta x=66^{\circ}+\frac{360^{\circ}}{\lambda} \times \frac{\lambda}{4}=66^{\circ}+90=156^{\circ}$
View full question & answer→MCQ 2431 Mark
White light may be considered to be a mixture of waves with $\lambda$ ranging between $3900 \mathring A$ and $7800 \mathring A$. An oil film of thickness 10,000 $\mathring A$ is examined normally by reflected light. If $\mu=1.4$, then the film appears bright for
- ✓
$4308 \mathring A, 5091 \mathring A, 6222 \mathring A$
- B
$4000 \mathring A, 5091 \mathring A, 5600 \mathring A$
- C
$4667 \mathring A, 6222 \mathring A, 7000 \mathring A$
- D
$4000 \mathring A, 4667 \mathring A, 5600 \mathring A, 7000 \mathring A$
AnswerCorrect option: A. $4308 \mathring A, 5091 \mathring A, 6222 \mathring A$
The film appears bright when the path difference $(2 \mu t \cos r)$ is equal to odd multiple of $\frac{\lambda}{2}$
i.e. $2 \mu t \cos r=(2 n-1) \lambda / 2 \quad$ where $n=1,2,3 \ldots$.
$\therefore \lambda=\frac{4 \mu t \cos r}{(2 n-1)}$
$=\frac{4 \times 1.4 \times 10,000 \times 10^{-10} \times \cos 0}{(2 n-1)}=\frac{56000}{(2 n-1)} \mathring A $
$\therefore \lambda=56000\ \mathring A ,18666\ \mathring A, 8000\ \mathring A, 6222\ \mathring A, 5091\ \mathring A, $
$4308\ \mathring A, 3733\ \mathring A .$
The wavelength which are not within specified range are to be refracted.
View full question & answer→MCQ 2441 Mark
light is incident normally on a diffraction grating through which the first order diffraction is seen at $32.$ The second order diffraction will be seen at
- A
$ 48$
- B
$ 64$
- C
$80$
- ✓
There is no second order diffraction in this case
AnswerCorrect option: D. There is no second order diffraction in this case
For a grating, $(e+d) \sin \theta_n=n \lambda$
where $(e+d)=$ greating element
$\sin \theta_n=\frac{n \lambda}{(e+d)}$
For $n=1, \sin \theta_1=\frac{\lambda}{(e+d)}=\sin 32^{\circ}$
This is more than 0.5 . Now $\sin \theta_2$ will be more than $2 \times 0.5$, which is not possible.
View full question & answer→MCQ 2451 Mark
The ratio of intensities of consecutive maxima in the diffraction pattern due to a single slit is
AnswerCorrect option: C. $1: \frac{4}{9 \pi^2}: \frac{4}{25 \pi^2}$
$I=I_0\left[\frac{\sin \alpha}{\alpha}\right]^2$, where $\alpha=\frac{\phi}{2}$For $n^{\text {th }}$
secondary maxima $d \sin \theta=\left(\frac{2 n+1}{2}\right) \lambda$
$\Rightarrow \alpha=\frac{\phi}{2}=\frac{\pi}{\lambda}[d \sin \theta]=\left(\frac{2 n+1}{2}\right) \pi $
$\therefore I=I_0\left[\frac{\sin \left(\frac{2 n+1}{2}\right) \pi}{\left(\frac{2 n+1}{2}\right) \pi}\right]^2=\frac{I_0}{\left\{\frac{(2 n+1)}{2} \pi\right\}^2}$
So $I_0: I_1: I_2=I_0: \frac{4}{9 \pi^2} I_0: \frac{4}{25 \pi^2} I_0$ $=1: \frac{4}{9 \pi^2}: \frac{4}{25 \pi^2}$
View full question & answer→MCQ 2461 Mark
In a single slit diffraction experiment first minimum for red light $(660 nm )$ coincides with first maximum of some other wavelength $\lambda^{\prime}$. The value of $\lambda^{\prime}$ is
- ✓
$4400 \mathring A$
- B
$6600 \mathring A$
- C
$2000 \mathring A$
- D
$3500 \mathring A$
AnswerCorrect option: A. $4400 \mathring A$
In a single slit diffraction experiment, position of minima is given by
$d \sin \theta=n \lambda$
So for first minima of red $\sin \theta=1 \times\left(\frac{\lambda_R}{d}\right)$and as first maxima is midway between first and second minima, for wavelength $\lambda^{\prime}$,its position will be
$d \sin \theta^{\prime}=\frac{\lambda^{\prime}+2 \lambda^{\prime}}{2} \Rightarrow \sin \theta^{\prime}=\frac{3 \lambda^{\prime}}{2 d}$
According to given condition $\sin \theta=\sin \theta^{\prime}$
$\Rightarrow \lambda^{\prime}=\frac{2}{3} \lambda_R \text { so } \lambda^{\prime}=\frac{2}{3} \times 6600=440 nm =4400 \mathring A$
View full question & answer→MCQ 2471 Mark
Angular width of central maxima in the Fraunhoffer diffraction pattern of a slit is measured. The slit is illuminated by light of wavelength $6000 \mathring A$. When the slit is illuminated by light of another wavelength, the angular width decreases by $30 \%$. The wavelength of this light will be
- A
$6000 \mathring A$
- ✓
$4200 \mathring A$
- C
$3000 \mathring A$
- D
$1800 \mathring A$
AnswerCorrect option: B. $4200 \mathring A$
$\text { Angular width } \beta=\frac{2 \lambda}{d}$
$ \Rightarrow \beta \propto \lambda $
$\Rightarrow \frac{\beta_1}{\beta_2}=\frac{\lambda_1}{\lambda_2} $
$\Rightarrow \frac{\beta}{\frac{70}{100} \beta}=\frac{6000}{\lambda_2}$
$\Rightarrow \lambda_2=4200 \mathring A$
View full question & answer→MCQ 2481 Mark
In a single slit diftraction of light of wavelength $\lambda$ by a slit of width $e$, the size of the central maximum on a screen at a distance $b$ is
- ✓
(a) $2 b \lambda+e$
- B
(b) $\frac{2 b \lambda}{e}$
- C
(c) $\frac{2 b \lambda}{e}+e$
- D
(d) $\frac{2 b \lambda}{e}-e$
AnswerCorrect option: A. (a) $2 b \lambda+e$
(a) In a single slit diffraction experiment, position of minima is given by $d \sin \theta=n \lambda$So for first minima of red $\sin \theta=1 \times\left(\frac{\lambda_R}{d}\right)$and as first maxima is midway between first and second minima, for wavelength $\lambda^{\prime}$,its position will be$d \sin \theta^{\prime}=\frac{\lambda^{\prime}+2 \lambda^{\prime}}{2} \Rightarrow \sin \theta^{\prime}=\frac{3 \lambda^{\prime}}{2 d}$According to given condition $\sin \theta=\sin \theta^{\prime}$$\Rightarrow \lambda^{\prime}=\frac{2}{3} \lambda_R \text { so } \lambda^{\prime}=\frac{2}{3} \times 6600=440 nm =4400A$
View full question & answer→MCQ 2491 Mark
A circular disc is placed in front of a narrow source. When the point of observation is $2 m$ from the disc, then it covers first HPZ. The intensity at this point is $l$. When the point of observation is $25 cm$ from the disc then intensity will be
- A
$\left(\frac{R_6}{R_2}\right)^2 I$
- B
$\left(\frac{R_7}{R_2}\right)^2 I$
- C
$\left(\frac{R_8}{R_2}\right)^2 I$
- ✓
$\left(\frac{R_9}{R_2}\right)^2 I$
AnswerCorrect option: D. $\left(\frac{R_9}{R_2}\right)^2 I$
$I=\frac{R_2^2}{4} \text { given } n_1 b_1=n_2 b_2$
$ \Rightarrow 1 \times 200=n_2 \times 25$
$\therefore n_2=8 H P Z $
$\therefore I=\left(\frac{R_9}{2}\right)^2 $
$=\left(\frac{R_9}{R_8} \times \frac{R_8}{R_7} \times \frac{R_7}{R_6} \times \frac{R_6}{R_5} \times \frac{R_5}{R_4} \times \frac{R_4}{R_3} \times \frac{R_3}{R_2} \times \frac{R_2}{R_2}\right)^2 $
$=\left(\frac{R_9}{R_2}\right)^2 I $
View full question & answer→MCQ 2501 Mark
A wavefront presents one, two and three HPZ at points $A, B$ and $C$ respectively. If the ratio of consecutive amplitudes of HPZ is $4: 3$, then the ratio of resultant intensities at these point will be
- A
$169: 16: 256$
- ✓
$256: 16: 169$
- C
$256: 16: 196$
- D
$256: 196: 16$
AnswerCorrect option: B. $256: 16: 169$
$I_A=R_1^2 $
$I_B=\left(R_1-R_2\right)^2=R_1^2\left(1-\frac{R_2}{R_1}\right)^2=R_1^2\left(1-\frac{3}{4}\right)^2=\frac{R_1^2}{16} $
$I_C=\left(R_1-R_2+R_3\right)^2=R_1^2\left(1-\frac{R_2}{R_1}+\frac{R_3}{R_1}\right)^2 $
$=R_1^2\left(1-\frac{R_2}{R_1}+\frac{R_3}{R_2} \times \frac{R_2}{R_1}\right)^2 \\=R_1^2\left(1-\frac{3}{4}+\frac{3}{4} \times \frac{3}{4}\right)^2=\left(\frac{13}{16}\right)^2 R_1^2=\frac{169}{256} R_1^2 $
$\therefore I_A: I_B: I_C=R_1^2: \frac{R_1^2}{16}: \frac{169}{256} R_1^2=256: 16: 169$
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