Since, the given lens is a convex lens hence it should converge a parallel beam of light falling on it to a single focal point. However, since it is showing the exactly opposite behaviour, thus it is definitely placed in a medium of higher refractive index.Thus, we conclude that \$ {\mu}_{1} > {\mu} \$

A ray of light is incident on the hypotenuse of a right-angled prism after travelling parallel to the base inside the prism. If \$\mu \$ is the refractive index of the material of the prism, the maximum value of the base angle for which light is totally reflected from the hypotenuse is :

\$\textbf {Hint :}\$ Use critical angle \$C=\sin^{-1}\left(\dfrac{1}{\mu}\right)\$\$\textbf{Step 1: }\$ Calculating the maximum value of the base angle \$\alpha\$If \$\alpha\$ is maximum value of base angle for which light is totally reflected from the hypotenuse and \$\theta\$ is the minimum value of angle of incidence at hypotenuse for total internal reflection.Then ,\$ \theta =sin^{-1}\left ( \dfrac{1}{\mu}\right )\$From diagram\$\alpha +\theta =90^o\$\$\Rightarrow \alpha = 90^o-sin^{-1}\left ( 1/\mu \right )=cos^{-1}\left ( 1/\mu \right )\$\${\textbf{Correct option: D}}\$

A beam of parallel rays is incident on a transparent slab of refractive index \$\sqrt{3}\$ making an angle \$30^{\circ}\$ with the surface of the slab. If the width of incident beam of light is 1.732 mm, the width of refracted beam is:

Using \$\mu _{1}sini=\mu _{2}sin\ r\$\$1\times sin60^{\circ}=\sqrt{3}\ sin\ r\$\$sin\ r=\dfrac{1}{2}\$\$AB=\dfrac{1.732mm}{cosi}\$Now,\$BB^{1}=AB\ cos\ r\$\$=\dfrac{1.732}{cosi}\times cos\ r\$\$=\dfrac{1.732}{\dfrac{1}{2}}\ \times \dfrac{\sqrt{3}}{2}=3\ mm\$

The refractive index of the material of a concave lens is \$\mu \$ and is placed in the medium of refractive index \$\mu ^{1}\$. If \$\mu ^{1}>\mu \$, then which of the following ray diagram is correct:

Since, the tendency of a concave lens is to diverge a parallel beam of light falling on the lens surface, it would show exactly the opposite behaviour if placed in a medium of higher refractive index than the material of the lens.Thus, the correct option will be one in which a concave lens diverges a parallel beam of a light ray.

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Ray Optics And Optical Instruments

Q: The difference in the number of wavelengths, when yellow light propagates through air and vacuum columns of the same thickness is one. The thickness of the air column is :(Refractive index of air \$\mu _{a}=\$1.0003 ; Wavelength of yellow light in vacuum \$= 6000A^{\circ}\$)

No of wavelengths \$=\dfrac{thickness}{wavelength}\$Let \$x\$ be the thickness of the air columnso \$\dfrac{x}{\lambda_{air}}-\dfrac{x}{\lambda _{vacuum}}=1\$\$\dfrac{x}{\lambda }\left ( 1.0003-1 \right )=1\$\$x=\dfrac{6000\times 10^{-10}}{0.0003}m\$\$=2 mm\$

Q: A ray of light is incident on a glass slab making an angle of \$60^0\$ with the surface. Calculate the angle of refraction in glass and the velocity of light in glass if the refractive index of glass and the velocity of light in air are 1.5 and 3.0 x 10\$^{8}\$ m/s respectively.

Using Snell's law :\$\mu_{1}sin\ i=\mu_{2}sin\ r\$\$1\times sin60^{\circ}=1.5\times sin\ r\$\$r=35^{\circ}15^{'}\$And \$V_{glass}=\dfrac {c}{\mu}\$\$=\dfrac {3\times 10^{8}}{1.5}\ m/s\$\$=2\times 10^{8}\ m/s\$

Q: In the case of refraction of light :a) Frequency changes b) Speed changes c) Wavelength changes

In the process of refraction, speed of light and wavelength of light changes. In a medium, speed of light is given by \$v=\dfrac{c}{\mu }\$ where \$\mu \$ is refractive index of the medium. Hence, for medium with different refractive index , speed of light is different. Frequency will not change because at the boundary/interface of the medium, the number of waves you send is the number of waves you receive at the other side, almost instantly. Since, for a travelling wave we have a relation \$ \nu = \dfrac{v}{\lambda} \$. Hence wavelength also changes in the process of refraction.So, option B is correct.

Q: A glass prism of refractive index \$1.5\$ is placed in water of refractive index \$1.33\$. The minimum value of the angle of the prism so that it will not be possible to have any emergent ray is :

\$\mu_{prism} =\$ \$1.5\$ \$\mu_{water}\$ \$=1.33\$\$\mu_{r} =\dfrac{1.5}{1.33}\$\$\theta _{cric} = sin^{-1} \left ( \dfrac{1}{\mu_{r}} \right )=sin^{-1}\left ( \dfrac{1.33}{1.5} \right )=62.5^{\circ}\$Maximum possible angle A is given by: \$A=2\theta _{cric}=125^{\circ}\$

Q: Two equi-convex lenses each of focal lengths 20 cm and refractive index 1.5 are placed in contact and space between them is filled with water of refractive index 4/3. The combination works as :

\$\begin{array}{l}\text { Ths is a combination of convex, concave and convex } \\\text { lens. Number them } 1,2,3 \\\qquad \begin{aligned}\text { given } f_{1} &=f_{3}=20 \mathrm{~cm} \\\mu_{1} &=\mu_{3}=1.5 \\\mu_{2} &=4/3\end{aligned}\end{array}\$\$\begin{array}{l}\quad \mu_{2}=4 / 3 \\\text { Now densmaker formula } \frac{1}{f}=\left(\mu_{r}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \\\text { for (1) } \quad \frac{1}{f_{1}}=(1.5-1)\left(\frac{1}{R_{1}}-\left(-\frac{1}{R}\right)\right)\end{array}\$\$\text { (equi convex lenS }\left.R_{1}=R_{2}\right)\$\$\begin{array}{l}\frac{1}{20}=\frac{1}{2}\times \frac{2}{R} \\ R=20 \mathrm{~cm} \\\text { for (2) } \\\qquad \begin{array}{l}\frac{1}{f_{2}}=\left(\frac{4}{3}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \\\frac{1}{f_{2}}=\frac{1}{3}\left(\frac{-2}{R}\right) \\ f_{2}=-30\mathrm{~cm}\end{array}\end{array}\$\$\begin{array}{l}\text { Now we know that } \\\qquad \begin{aligned}\text { ieffective } &=P_{1}+P_{2}+P_{3} \\\frac{1}{f_{e q}} &=\frac{1}{20}-\frac{1}{30}+\frac{1}{20} \\ f_{\text {eq }}=15\mathrm{~cm}\end{aligned} \\\text { So it is a converging lens of focal length } 15 \mathrm{~cm} .\end{array} \$

Q: A ray incident at a point at an angle of incidence \$60\$\$^{\circ}\$ enters a glass sphere of \$\mu =\sqrt{3}\$ and is reflected and refracted at the further surface of the sphere. The angle between the reflected and refracted rays at this surface is :

Using Snell's Law at point \$P\$ we get:-\$\dfrac{sin\ 60^o}{sin\ r_1}=\sqrt3\$\$sin \ r_1= \dfrac12\implies r_1= 30^o\$As we know that \$r_1=r_2 \implies r_2=30^o\$Using Snell's Law at point \$Q\$ we get:-\$\dfrac{sin \ 30^o}{sin\ i_2}=\dfrac1{\sqrt3}\implies i_2= 60^o\$Since reflection at point \$Q\$ occurs\$\implies r_2'=r_2= 60^o\$Let angle between the refracted ray and reflected ray be \$\alpha\$\$\alpha= 180^o- (r_2'+r_2)= 90^o\$Hence option \$(B)\$ is correct.

Physics
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Easy Qs Med Qs Hard Qs

The difference in the number of wavelengths, when yellow light propagates through air and vacuum columns of the same thickness is one. The thickness of the air column is :
(Refractive index of air $μ_{a} =$ 1.0003 ; Wavelength of yellow light in vacuum $= 6000 A^{\circ}$ )

1.8 m m

2 m m

2 c m

2.2 c m

Medium

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A ray of light is incident on a glass slab making an angle of $6 0^{0}$ with the surface. Calculate the angle of refraction in glass and the velocity of light in glass if the refractive index of glass and the velocity of light in air are 1.5 and 3.0 x 10 $^{8}$ m/s respectively.

2 8^{\circ} 3 0^{'}

3 \times 10

^{8}

^{- 1}

4 0^{\circ} 2 0^{'}, 2.5 \times 10

^{8}

^{- 1}

3 5^{\circ} 1 5^{'}, 2 \times 1 0^{8}

^{- 1}

2 8^{\circ} 3 0^{'}, 2.5 \times 1 0^{8}

^{- 1}

Medium

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In the case of refraction of light :
a) Frequency changes
b) Speed changes
c) Wavelength changes

a is correct

b and c are correct

a, b, c are correct

a and b are correct

Medium

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In the diagram shown

μ^{1} = μ

μ^{1} < μ

μ^{1} > μ

μ^{1} \geq μ

Medium

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A ray of light is incident on the hypotenuse of a right-angled prism after travelling parallel to the base inside the prism. If $μ$ is the refractive index of the material of the prism, the maximum value of the base angle for which light is totally reflected from the hypotenuse is :

s i n^{- 1} (\frac{1}{μ})

t a n^{- 1} (\frac{1}{μ})

s i n^{- 1} (\frac{μ - 1}{μ})

c o s^{- 1} (\frac{1}{μ})

Medium

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A beam of parallel rays is incident on a transparent slab of refractive index $3$ making an angle $3 0^{\circ}$ with the surface of the slab. If the width of incident beam of light is 1.732 mm, the width of refracted beam is:

1.00 mm

1.50mm

2.50 mm

3.00 mm

Medium

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A glass prism of refractive index $1.5$ is placed in water of refractive index $1.33$ . The minimum value of the angle of the prism so that it will not be possible to have any emergent ray is :

15 0^{\circ}

12 5^{\circ}

16 5^{\circ}

18 0^{\circ}

Medium

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Two equi-convex lenses each of focal lengths 20 cm and refractive index 1.5 are placed in contact and space between them is filled with water of refractive index 4/3. The combination works as :

converging lens of focal length 50 cm.

diverging lens of focal length 15 cm

converging lens of focal length 15 cm

diverging lens of focal length 50 cm

Medium

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The refractive index of the material of a concave lens is $μ$ and is placed in the medium of refractive index $μ^{1}$ . If $μ^{1} > μ$ , then which of the following ray diagram is correct:

None of these

Medium

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A ray incident at a point at an angle of incidence $60$ $^{\circ}$ enters a glass sphere of $μ = 3$ and is reflected and refracted at the further surface of the sphere. The angle between the reflected and refracted rays at this surface is :

50

^{\circ}

90

^{\circ}

60

^{\circ}

40

^{\circ}

Medium

View solution>

Ray Optics And Optical Instruments

Physics 6735 Views

The difference in the number of wavelengths, when yellow light propagates through air and vacuum columns of the same thickness is one. The thickness of the air column is :(Refractive index of air μa​=1.0003 ; Wavelength of yellow light in vacuum =6000A∘)

A ray of light is incident on a glass slab making an angle of 600 with the surface. Calculate the angle of refraction in glass and the velocity of light in glass if the refractive index of glass and the velocity of light in air are 1.5 and 3.0 x 108 m/s respectively.

In the case of refraction of light : a) Frequency changes b) Speed changes c) Wavelength changes

In the diagram shown

A ray of light is incident on the hypotenuse of a right-angled prism after travelling parallel to the base inside the prism. If μ is the refractive index of the material of the prism, the maximum value of the base angle for which light is totally reflected from the hypotenuse is :

A beam of parallel rays is incident on a transparent slab of refractive index 3​ making an angle 30∘ with the surface of the slab. If the width of incident beam of light is 1.732 mm, the width of refracted beam is:

A glass prism of refractive index 1.5 is placed in water of refractive index 1.33. The minimum value of the angle of the prism so that it will not be possible to have any emergent ray is :

Two equi-convex lenses each of focal lengths 20 cm and refractive index 1.5 are placed in contact and space between them is filled with water of refractive index 4/3. The combination works as :

The refractive index of the material of a concave lens is μ and is placed in the medium of refractive index μ1. If μ1>μ, then which of the following ray diagram is correct:

A ray incident at a point at an angle of incidence 60∘ enters a glass sphere of μ=3​ and is reflected and refracted at the further surface of the sphere. The angle between the reflected and refracted rays at this surface is :

Physics
6735 Views

The difference in the number of wavelengths, when yellow light propagates through air and vacuum columns of the same thickness is one. The thickness of the air column is :
(Refractive index of air $μ_{a} =$ 1.0003 ; Wavelength of yellow light in vacuum $= 6000 A^{\circ}$ )

A ray of light is incident on a glass slab making an angle of $6 0^{0}$ with the surface. Calculate the angle of refraction in glass and the velocity of light in glass if the refractive index of glass and the velocity of light in air are 1.5 and 3.0 x 10 $^{8}$ m/s respectively.

In the case of refraction of light :
a) Frequency changes
b) Speed changes
c) Wavelength changes

A ray of light is incident on the hypotenuse of a right-angled prism after travelling parallel to the base inside the prism. If $μ$ is the refractive index of the material of the prism, the maximum value of the base angle for which light is totally reflected from the hypotenuse is :

A beam of parallel rays is incident on a transparent slab of refractive index $3$ making an angle $3 0^{\circ}$ with the surface of the slab. If the width of incident beam of light is 1.732 mm, the width of refracted beam is:

A glass prism of refractive index $1.5$ is placed in water of refractive index $1.33$ . The minimum value of the angle of the prism so that it will not be possible to have any emergent ray is :

The refractive index of the material of a concave lens is $μ$ and is placed in the medium of refractive index $μ^{1}$ . If $μ^{1} > μ$ , then which of the following ray diagram is correct:

A ray incident at a point at an angle of incidence $60$ $^{\circ}$ enters a glass sphere of $μ = 3$ and is reflected and refracted at the further surface of the sphere. The angle between the reflected and refracted rays at this surface is :