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Absolute RI of diamond= 1.6 

Absolute RI of glass= 1.5 

Multiplying them we get 

= 1.5 x 1.6 

 = 2.4

Refractive index of air, μair = 1; 

Given; 

Refractive index of water w.r.t air, aμw = \(\frac43\)

Refractive index of glass w.r.t air, aμg = \(\frac32\)

Refractive index of water w.r.t glass = gμw = \(\frac{^aμ_w}{^aμ_g}\,=\,\frac{\frac43}{\frac32}\)

⇒ gμw = \(\frac{4\times2}{3\times3}\,=\,\frac89\)

∴ Refractive index of water w.r.t glass is = \(\frac89\)

Given; 

Refractive index of alcohol (μ) = 1.36. 

Speed of light in vacuum = 3×10⁸m/s. 

Refractive index (μ) = \(\frac{c\,(speed\,of\,light\,in\,vaccum)}{v\,(speed\,of\,light\,in\,alcohol)}\)

⇒ 1.36 = \(\frac{3\,\times\,10^8}{speed\,of\,light\,in\,alcohol}\)

⇒ speed of light in alcohol = \(\frac{3\,\times\,10^8}{1.36}\)

∴ Speed of light in alcohol = 2.21 × 10⁸m/s 

Speed of light in alcohol with respect to air is 2.21 x 10⁸ms^-1

Refractive index of air, μair = 1; 

Given; 

Refractive index of medium x w.r.t medium y, μxy = \(\frac23\)

Refractive index of medium y w.r.t medium z, μyz = \(\frac43\)

Refractive index of medium z w.r.t medium x = μzx = \(\frac{μ_z}{μ_x} = \frac{μ_z}{μ_y}\times\frac{μ_y}{μ_x}\)

⇒ μzx = \(\frac{1}{\frac{μ_y}{μ_z}}\times\frac{1}{\frac{μ_x}{μ_y}}\)

⇒ μzx = \(\frac{1}{μ_{yz}}\times\frac{1}{μ_{xy}}\)

⇒ μzx = \(\frac{1}{\frac{4}{3}}\times\frac{1}{\frac{2}{3}}\)

⇒ μzx = \(\frac{3}{4}\times\frac32\)

⇒ μzx = \(\frac98\)

∴ Refractive index of medium z w.r.t medium x is .\(\frac98\)

(i) Image formed by plane mirror is of the same size as of the object. Positive value of magnification indicates that image formed by plane mirror is erect. Image formed by plane mirror is at same distance as the object but behind the mirror. Magnification of image is +1.

 The image is virtual. 

(ii) Varun made an arrangement for the light so that his mother could finish the work easily this shows that Varun is caring and responsible towards his mother. Also, he has used his knowledge of plane mirrors to get the light this shows that he is also quite clever and intelligent also he is the serious student who focuses on the practical aspect of things taught in class and applies them in a day to day use. 

In other words, Varun had put the science to the real use.

When the speed of light is more as compared to other medium.

The degree of convergence or divergence of light rays achieved by a lens is expressed in terms of its power.

When the speed of light is less as compared to another medium.

A medium in which light travels comparatively slower than the other medium is called an optically denser medium.

Correct option is (B) speed of light remains constant.

A ray of light enters from kerosene oil to water i.e., refractive index 1.44 to 1.33 i.e., from denser to rarer medium. Hence the ray of light bends away from the normal.

1. Snell’s law is defined as “The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, for the light of a given colour and for the given pair of media”. Snell’s law formula is expressed as:

sin i / sin r = constant = μ

Where i is the angle of incidence and r is the angle of refraction. This constant value is called the refractive index of the second medium with respect to the first.

2. The point on the spherical surface of mirror through which principal axis passes is called as pole of the concave mirror.

1ⁿ2 = \(\frac{n_2}{n_1}\) and

2ⁿ1 = \(\frac{n_1}{n_2}\)

⇒ 1ⁿ2 x 2ⁿ1 = \(\frac{n_2}{n_1}\) x \(\frac{n_1}{n_2}\) = 1.

Hence the answer is 1.

Refractive index is the measure of bending of ray of light as it passes from one medium to another. It is also defined as the ratio of the speed of light in vacuum to the speed of light in any medium.

Refractive index (μ) = \(\frac{c\,(spedd\,of\,light\,in\,vaccum)}{v\,(speed\,of\,light\,in\,any\,medium)}\)

Given

Speed of light in a medium (v) =\(\frac{2c}{3}\)

⇒ µ = \(\frac{c}{\frac{2c}{2}}\)

⇒ μ =\(\frac{3c}{2c}\)

⇒ μ = \(\frac32\)= 1.5

Hence Refractive index of medium is 1.5.

Given; 

Refractive index of (R.I) water w.r.t vacuum = \(\frac43\)

Refractive index (R.I) of vacuum w.r.t glass = \(\frac23\)

Refractive index (R.I) of glass to vacuum will be = \(\frac32\)

Refractive index of glass to water = \(\frac{R.I\,of\,glass}{R.I\,of\,water}= \frac{\frac32}{\frac43}\)

⇒ Refractive index of glass w.r.t water = \(\frac{3\times3}{4\times2}= \frac98\)

⇒ \(\frac{speed\,of\,light\,in\,water}{speed\,of\,light\,in\,glass}= \frac98\)

⇒ \(\frac{speed\,of\,light\,in\,water}{2\times10^8}= \frac98\)

⇒ Speed of light in water = \(\frac{9\times2\times10^8}{8}\)

⇒ Speed of light in water = \(\frac{9\times10^8}{4}\) = 2.25 x 10⁸

Thus, speed of light in water = 2.25 × 10⁸

Refractive index of water w.r.t vacuum = vμw = \(\frac43\)

⇒ \(\frac{speed\,of\,light\,in\,vacuum}{speed\,of\,light\,in\,water}= \frac43\)

⇒ \(\frac{speed\,of\,light\,in\,vacuum}{2.25\times10^8}= \frac43\)

 ⇒ Speed of light in water = \(\frac{4\times2.25\times10^8}{3}\)

⇒ Speed of light in water = \(\frac{9\times10^8}{3}\) 

Thus, speed of light in vacuum is 3 × 10⁸ m/s.

Let velocity of light in vacuum = v 

⇒ Velocity of light in water = \(\frac34\) v

Now,

Refractive index of a medium (water) = \(\frac{Speed\,of\,light\,in\,vacuum}{Speed\,of\,light\,in\,meium\,(water)}\)

⇒ Refractive index = \(\frac34\) = 1.33 ...

Hence Refractive index of water = 1.33.

(a) The values shown by Shiva are :

(i) presence of mind,

(ii) high degree of general awareness, and

(iii) social responsibility. 

(b) Plane, concave and convex mirrors. 

(c) The special mirror is the combination of three mirrors which show their characteristics property.

(i) Convex mirror is the upper portion of the mirror.

(ii) Concave mirror is the middle portion of the mirror.

(iii) Plane mirror is the lower portion of the mirror.

As we move above the earth surface the density of air goes on decreasing because the amount of air present at high altitudes is less. Thus due to the decrease in the density of the air, the refractive index gradually increases from the atmosphere to the earth’s surface. Therefore the refractive index of the atmosphere is different at different altitudes.

(a) (i) The ray of light should travel from denser to rarer medium.

(ii) The angle of incidence should be more than the critical angle.

(b) μ = 1/sin i_C  

Where i_C is the critical angle.

we know that glass is denser medium than water. Therefore, a ray of light will bend towards the normal when it goes from water to glass.

A) X and Y are true

Optical centre.

A) Both statements are true

Fyokla felt that something big, heavy and round was rolling over the sky, and tearing it’ open exactly over her head.

The wood had broken at the edge of the hole and jammed Danilka’s hand.

Trinity said that it was not from spite that it thundered and it would not kill a little thing like Fyokla

The boy was gazing at the story sky and apparently he forgot to think of his trouble.

Trinity was a tall old man with a thin, pockmarked face, very long leg, and bare feet, dressed in a woman’s tattered jacket. On his long crane-like legs, he swayed in the wind.

The boy does not sleep. He gazes into the darkness and it seems to him that he can see everything that he has seen during the day.

He knows the names of all the wildflowers, animals and stones. He knows the herbs that cure diseases, he has no difficulty in telling the age of a horse or a cow. Looking at the sunset, the moon, or the birds, he could tell what sort of weather it would be the next day.

Anton Chekhov: 

Anton Chekov was born on January 29, 1860, in Russia. He was famous as a physician and writer. He was awarded the Pushkin Prize. The Steppe, The Lady with the dog, A Day in the Country, etc are his notable works. Chekov died of tuberculosis on July 15, 1904, in Germany.

My father/mother/teacher/friend/M.T Vasudevan Nair/Sachin Tendulkar/Gandhiji/etc. You can also get inspiration from a desire to be the best you can be.

The expression, “I slept fitfully”.

She was quite honest.

She was listening for human voices. But what she could hear was the screaming of monkeys and the screeching of parakeets. So she was disappointed.

The attack might lead to the amputation of her limbs.

• I would consider each challenge as a blessing to know more about life. 
• I would look for ways to overcome the adverse situation. 
• I would realize that I still have options and then assert exactly what I want for myself. 
• I would pay greater attention to what is going right for me than what is not. With hope, determination, and perseverance, 
• I would try to overcome the disabilities, disasters or accidents. 
• I would believe in what Charles Swindoll told, “Life is 10% what happens to you and 90% how you react to it.” 
• I would believe that every difficult situation will enrich my life – the life that will be and the life I have right now.

Success is no accident. It is hard work, perseverance, learning, studying, sacrifice and most of all, love of what you are doing – Pele.

(1) Stream line flow : Stream line flow of a liquid is that flow in which each element of the liquid passing through a point travels along the same path and with the same velocity as the preceding element passes through that point. 

The two streamlines cannot cross each other and the greater is the crowding of streamlines at a place, the greater is the velocity of liquid particles at that place. 

(2) Laminar flow : If a liquid is flowing a horizontal surface with a steady flow and moves in the form of layers of different velocities which do not mix with each other, then the flow of liquid is called laminar flow. 

In this flow the velocity of liquid flow is always less than the critical velocity of the liquid. 

(3) Turbulent flow : When a liquid moves with a velocity greater than its critical velocity, the motion of the particles of liquid becomes disordered or irregular. Such a flow is called a turbulent flow. 

The critical velocity is that velocity of liquid flow upto which its flow is streamlined and above which its flow becomes turbulent.

Given \(\lambda\) = 5pm = 5 x 10^-12 m

Velocity of photon = Velocity of light= 3.0 × 10⁸m/sec.

\(\therefore\) \(\lambda\) = \(\frac{h}{mv}\)

or m = \(\frac{h}{\lambda v}\) = \(\frac{6.626\times10^{-34}Js}{5\times10^{-12}m\times3.0\times10^8m/sec}\)

= 4.417 × 10⁻²⁷Kg

For a diatomic gas,
C_P = \(\frac { 7 }{ 2 } \) R, where R = Universal gas constant

An airplane can have independent velocities in three directions (X,Y,Z). Hence, the degree of freedom is 3.

For any system in thermal equilibrium, the total energy is equally distributed among its various degree of freedom. And the energy associated with each molecule of the system per degree of freedom of the system is 1/2 kT.

Correct option is B) 1 J/s

Correct option: (b) boil 

In this case, weight of the stone is balanced by the normal reaction. The net force on the stone is given by:

F = Ma = 0.1 x 1 = 0.1 N (along horizontal)

In the train, accelerating with 1 ms^-2, the stone experiences on additional force F (also horizontal) given by:

F' = Ma = 0.1 x 1 = 0.1 N

As the stone is dropped, the force F' no longer acts and the net force acting on the stone is,

F' = Mg = 0.98 N (vertically downwards)

When the stone is dropped from the window of a train running at a constant velocity. no force acts on the stone due to motion of the train.

Thus, Force on stone = Mg - 0.98 N (vertically downward)

Here, g = 9.8 ms^-2. When stone of mass 0.1 kg is dropped from the window of a stationary train, the net force acting on it just after it is dropped, will be

F = weight of stone = Mg = 0.1 x 9.8 N

= 0.98 N (vertically downward)