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In an experiment for determination of refractive index of glass of a prism by $i\, -\, \delta$ , plot, it was found that a ray incident at angle 35⁰, suffers a deviation of 40⁰ and that it emerges at angle 79⁰⋅ Ιn that case which of the following is closest to the maximum possible value of the refractive index ?
Option: 1 1.5
Option: 2 1.6
Option: 3 1.7
Option: 4 1.8

This angle is greater than the 40° deviation angle already given, For greater μ, deviation will be even higher. Hence, μ of the given prism should be lesser than 1.5. Hence, the closest option will be 1.5.

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Posted by

Ritika Jonwal

A single slit of width b is illuminated by a coherent monochromatic light of wavelength λ. If the second and fourth minima in the diffraction pattern at a distance 1 m from the slit are at 3 cm and 6 cm respectively from the central maximum, what is the width of the central maximum ? (i.e. distance between first minimum on either side of the central maximum)
Option: 1 1.5 cm
Option: 2 3.0 cm
Option: 3 4.5 cm
Option: 4 6.0 cm

As the distance of nth minima from the central maximum is given as

$x_{n}=\frac{D}{d}\left[n \frac{\lambda}{2}\right]$

Difference between second and fourth minimum =6−3=3

i.e $x_4-x_2=3$

$x_4=\frac{4\lambda D}{2d} \ \ and \ \ x_2=\frac{2 \lambda D}{2d} \\ \ so \ \ x_4-x_2=3=\frac{2\lambda D}{2d} \\ \Rightarrow \frac{\lambda D}{d}=3 \ cm$

$\text {Now, width of central maximum }=2 \times x_{1}$ $=2 \times \frac{D}{d}\left[1* \frac{\lambda}{2}\right]$ $=\frac{D \lambda}{d}$ $=3 \mathrm{cm}$

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Posted by

vishal kumar

A point object in air is in front of the curved surface of a $plano-convex$ lens. The radius of curvature of the curved surface is $30cm$ and the refractive index of the lens material is $1.5,$ then the focal length of the lens (in cm) is _____________.
Option: 1 60
Option: 2 30
Option: 3 120
Option: 4 100

So, $\frac{1}{f}=(1.5-1)\left(\frac{1}{30}-\frac{1}{\infty}\right)=\frac{0.5}{30}=\frac{1}{60}$

So f= 60 cm

So the answer will be 60.

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Posted by

vishal kumar

The magnifying power of a telescope with tube length $60cm$ is $5$ . What is the focal length (in cm) of its eye piece?
Option: 1 30
Option: 2 40
Option: 3 10
Option: 4 20

$\begin{array}{l}{\text { Let the focal length of the objective as } f_{0}} \\ {\text { and focal length of the eyepiece as } f_{e}} \\ {\text { Magnifying power, } M=\frac{f_{0}}{f_{e}}} \\ {\text { Tube length, } L=f_{0}+f_{e}}\end{array}$

$\begin{array}{l}{\text { Given: Magnifying power }=5} \\ \\ {\therefore \frac{f_{0}}{f_{\mathrm{e}}}=5} \\ {\therefore {f}_{0}=5{f}_{e}} \\ and \ {\mathrm{L}=\mathrm{f}_{0}+\mathrm{f}_{\mathrm{e}}=60} \\ {5 \mathrm{f}_{\mathrm{e}}+\mathrm{f}_{\mathrm{e}}=60\quad\left[\because \mathrm{f}_{0}=5 \mathrm{f}_{\mathrm{e}}\right]} \\ {6\mathrm{f}_{\mathrm{e}}=60} \\ {\mathrm{f}_{\mathrm{e}}=\frac{60}{6}=10 \mathrm{cm}} \\ {\text { So the focal length of eye piece is } 10 \mathrm{cm} .}\end{array}$

So the correct answer is given in option 3.

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Posted by

vishal kumar

The critical angle (in degree)of a medium for a specific wavelength, if the medium has relative permitivity $3$ and relative permeability $\frac{4}{3}$ for this wavelength, will be :
Option: 1 15
Option: 2 30
Option: 3 45
Option: 4 60

$C_{vacuum}=\frac{1}{\sqrt{\mu _0\epsilon _0}} \ and \ C_{medium}=\frac{1}{\sqrt{\mu _0\mu _r\epsilon _0\epsilon _r}}$

and $n=\frac{C_{vacuum}}{C_{medium}}=\sqrt{\mu _r\epsilon _r} =\sqrt{\frac{4}{3}*3}=2$

By using snells law

$n*sinC=1*sin(90^0)\\ \Rightarrow n*sinC=1\\ \Rightarrow sinC=\frac{1}{2}\Rightarrow critical \ angle =C=30^0$

So the correct option is 2.

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Posted by

vishal kumar

In a young's double slit experiment 15 fringes are observed on a small portion of the screen when light of wavelength 500 nm is used. Ten fringes are observed on the same section of the screen when another light source of wavelength $\lambda$ is used. Then the value of $\lambda$ is ( in nm) _____.
Option: 1 750
Option: 2 650
Option: 3 850
Option: 4 450

$15\times 500\times \frac{D}{d}=10\times \lambda _{2}\times \frac{D}{d}$

$\lambda _{2}=15\times 50nm$

$\lambda _{2}=750nm$

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Posted by

avinash.dongre

Visible height of wavelength $6000\times10^{-8}$ cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction min. is at 60 from the central max. If the first minimum is produced at $\theta_{1}$ , the $\theta_{1}$ is close to,

Option: 1 $45^{o}$
Option: 2 $30^{o}$
Option: 3 $25^{o}$
Option: 4 $20^{o}$

Fraunhofer diffraction by a single slit -

Fraunhofer diffraction by a single slit

let's assume a plane wave front is incident on a slit AB (of width b). Each and every part of the exposed part of the plane wave front (i.e. every part of the slit) acts as a source of secondary wavelets spreading in all directions. The diffraction is obtained on a screen placed at a large distance. (In practice, this condition is achieved by placing the screen at the focal plane of a converging lens placed just after the slit).

The diffraction pattern consists of a central bright fringe (central maxima) surrounded by dark and bright lines (called secondary minima and maxima).
At point O on the screen, the central maxima is obtained. The wavelets originating from points A and B meets in the same phase at this point, hence at O, intensity is maximum

Secondary minima : For obtaining nth secondary minima at P on the screen, path difference between the diffracted waves

$\Delta x =b\sin{\theta }=n \lambda$

Angular position of nth secondary minima:

$\sin{\theta } \approx \theta = \frac{n\lambda} {b}$

2. Distance of nth secondary minima from central maxima:

$x_{n}=D\cdot \theta =\frac{n\lambda D}{b}$ where D = Distance between slit and screen.

$f\approx D$ = Focal length of converging lens.

Secondary maxima : For nth secondary maxima at P on the screen.

Path difference $\Delta x =b\sin{\theta }=(2n+1) \frac{\lambda}{2}$ ; where n = 1, 2, 3 .....

(i) Angular position of nth secondary maxima

$\sin{\theta } \approx \theta \approx \frac{(2n+1)\lambda} {2b}$

(ii) Distance of nth secondary maxima from central maxima:

$x_{n}=D\cdot \theta =\frac{(2n+1)\lambda D}{2b}$

Central maxima : The central maxima lies between the first minima on both sides.

(i) The Angular width d central maxima = $2 \theta=\frac{2 \lambda}{b}$
(ii) Linear width of central maxima $=2 x=2 D \theta=2 f \theta=\frac{2 \lambda f}{b}$

Intensity distribution: if the intensity of the central maxima is $I_{0}$ then the intensity of the first and secondary maxima are
found to be $\frac{I_{0}}{22}$ and $\frac{I_{0}}{61}$ . Thus diffraction fringes are of unequal width and unequal intenstities.

(i) The mathematical expression for in intensity distribution on the screen is given by:

$I=I_{o}\left(\frac{\sin \alpha}{\alpha}\right)^{2}$ where $\alpha$ is just a convenient connection between the angle $\theta$ that locates a point on the viewing screening and light intensity I.

$\phi=$ Phase difference between the top and bottom ray from the slit width b.
Also $\alpha=\frac{1}{2} \phi=\frac{\pi b}{\lambda} \sin \theta$ .

(ii) As the slit width increases relative to wavelength the width of the control diffraction maxima decreases; that is, the light undergoes less flaring by the slit. The secondary maxima also decreases in width and becomes weaker.

(iii) If b>>λ, the secondary maxima due to the slit disappear; we then no longer have single slit diffraction.

-

$d\ sin\theta=path\ difference=2\lambda$

So,

$\\d\ sin\60=2\lambda\\\Rightarrow \frac{\lambda}{d}=\frac{\sqrt{3}}{4}$

For first minima,

$\\d\ sin\theta_2=\lambda\\sin\theta_2=\frac{\lambda}{d}=\frac{\sqrt{3}}{4}\\\Rightarrow \theta_2=sin^{-1}(\frac{\sqrt{3}}{4})=25.64^\circ=25^\circ$

So option (3) is correct.

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Posted by

Ritika Jonwal

If we used a magnification of 375 from a compound microscope of tube length 150mm and an objective of focal length 5mm, the focal length of the eyepiece should be close to :

Option: 1 12mm
Option: 233mm
Option: 3 22mm
Option: 4 2mm

The magnification is given by

$M=\frac{L}{f_0}(1+\frac{D}{f_e})\\ \Rightarrow 375=\frac{150}{5}(1+\frac{25}{f_e})\\ \Rightarrow f_e=22 \ mm$

So option (3) is correct.

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Posted by

Ritika Jonwal

There is a small source of light at some depth below the surface of water (refractive index = $\frac{4}{3}$ ) in a tank of large cross sectional surface area. Neglecting any reflection from the bottom and absorption by water, percentage of light that emerges out of surface is (nearly) : [ Use the fact that surface area of a spherical cap of height h and radius of curvature r is $2\pi rh$ ]
Option: 1 $34\; ^{o}/_{o}$
Option: 2 $17\; ^{o}/_{o}$
Option: 3 $50\; ^{o}/_{o}$
Option: 4 $21\; ^{o}/_{o}$

$\sin \beta=\frac{3}{4},\cos \beta =\frac{\sqrt{7}}{4}$

$Solid\; angle\; d\Omega =2\pi R^{2}(1-\cos \beta )$

Percentage of light = $\frac{2\pi R^{2}(1-\cos \beta )}{4\pi R^{2}}\times 100=\frac{1-\cos \beta }{2}\times 100=\left ( \frac{4-\sqrt{7}}{8} \right )\times 100\approx 17\: ^{o}/_{o}$

Hence the correct option is (2).

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Posted by

avinash.dongre

In a double-slit experiment, at a certain point on the screen the path difference betweeen the two interfering waves is $\frac{1}{8}$ th of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe is :
Option: 1 0.568
Option: 2 0.853
Option: 3 0.760
Option: 4 0.672

Intensity is given by $I=I_{0} \cos ^{2}\left(\frac{\Delta \phi}{2}\right)$

Where I_O=maximum intensity=intensity of light at the center of a bright fringe

and using the Relation $\Delta x=\frac{2\pi }{\lambda }\Delta \phi$

we get $\frac{I}{\mathrm{I}_{0}}=\cos ^{2}\left[\frac{\frac{2 \pi}{\lambda} \times \Delta x}{2}\right]=\cos ^{2}\left(\frac{\pi}{8}\right) ; \quad \frac{I}{\mathrm{I}_{0}}=0.853$

Hence the correct option is (2).

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Posted by

Ritika Jonwal

Posted by

vishal kumar

Crack CUET with india's "Best Teachers"

A point object in air is in front of the curved surface of a lens. The radius of curvature of the curved surface is and the refractive index of the lens material is then the focal length of the lens (in cm) is _____________. Option: 1 60 Option: 2 30 Option: 3 120 Option: 4 100

Posted by

vishal kumar

The magnifying power of a telescope with tube length is . What is the focal length (in cm) of its eye piece? Option: 1 30 Option: 2 40 Option: 3 10 Option: 4 20

Posted by

vishal kumar

JEE Main high-scoring chapters and topics

The critical angle (in degree)of a medium for a specific wavelength, if the medium has relative permitivity and relative permeability for this wavelength, will be : Option: 1 15 Option: 2 30 Option: 3 45 Option: 4 60

Posted by

vishal kumar

The magnifying power of a telescope with tube length $60cm$ is $5$ . What is the focal length (in cm) of its eye piece?
Option: 1 30
Option: 2 40
Option: 3 10
Option: 4 20

The critical angle (in degree)of a medium for a specific wavelength, if the medium has relative permitivity $3$ and relative permeability $\frac{4}{3}$ for this wavelength, will be :
Option: 1 15
Option: 2 30
Option: 3 45
Option: 4 60

If we used a magnification of 375 from a compound microscope of tube length 150mm and an objective of focal length 5mm, the focal length of the eyepiece should be close to :

Option: 1 12mm
Option: 233mm
Option: 3 22mm
Option: 4 2mm