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#### In an experiment for determination of refractive index of glass of a prism by , 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.

#### 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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#### 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

So,

So f= 60 cm

So the answer will be 60.

#### 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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#### 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

and

By using snells law

So the correct option is 2.

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#### Visible height of wavelength 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 , the  is close to,   Option: 1 Option: 2  Option: 3    Option: 4

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

1. Angular position of nth secondary minima:

2. Distance of nth secondary minima from central maxima:

where D = Distance between slit and screen.

= Focal length of converging lens.

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

Path difference      ; where n = 1, 2, 3 .....

(i) Angular position of nth secondary maxima

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

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

(i) The Angular width d central maxima    =
(ii) Linear width of central maxima

Intensity distribution: if the intensity of the central maxima is    then the intensity of the first and secondary maxima are
found to be   . Thus diffraction fringes are of unequal width and unequal intenstities.

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

where   is just a convenient connection between the angle that locates a point on the viewing screening and light intensity I.

Phase difference between the top and bottom ray from the slit width b.
Also

(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.

-

So,

For first minima,

So option (3) is correct.

#### 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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#### There is a small source of light at some depth below the surface of water (refractive index = ) 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  ] Option: 1       Option: 2       Option: 3       Option: 4

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

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

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).

#### In a double-slit experiment, at a certain point on the screen the path difference betweeen the two interfering waves is 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

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

and  using the Relation

we get

Hence the correct option is (2).