# AP SSC 10th Class Physical Science Chapter 5 Solution – Refraction of Light at Plane Surface

## AP SSC 10th Class Physical Science Chapter 5 Solution – Refraction of Light at Plane Surface

Andhra Pradesh SSC 10th Class Physical Science Chapter 5 Refraction of Light at Plane Surface Solution for AP SSC 10th Class Physics/Chemistry Exam. Lots of Students of Andhra Pradesh Board will search on internet for Andhra Pradesh Class 10 Physical Science Textbook Solution or Study Material for AP SSC 10th exam. Here you search will end! Here in this page we have provided for all question answer for Chapter 5 Refraction of Light at Plane Surface.

For any help regarding extra notes for AP SSC 10th Class Physics/Chemistry Chapter 5 Refraction of Light at Plane Surface you can comment us at comment box below.

### AP 10th Class Physics Chapter 5 Refraction of Light at Plane Surface Solution

(1) Why is it difficult to shoot a fish swimming in water? (AS1)

Ans: –The refraction of water and air interference made it difficult to shoot a fish swimming in water. For this reason, sometimes we see fish in the upper and sometimes in down causing it difficult to film.

(2) The speed of the light in a diamond is 1, 24, 000 km/s. Find the refractive index of diamond if the speed of light in air is 3,00,000 km/s. (AS1)

Ans: –The refractive index of diamond is = (speed of the light in air /speed of the light in diamond). Or, refractive index = 300000km/s / 124000km/s = 2.4 Then the refractive index of diamond is 2.4.

(3) Refractive index of glass relative to water is 9/8. What is the refractive index of water relative to glass? (AS1)

Ans: –By questions the refractive index of glass relative to water= 9/8 The refractive index of water relative to glass is = (1/refractive index glass relative to water). = (1/9/8) = 8/9.

(4) The absolute refractive index of water is 4/3. What is the critical angle? (AS1)

Ans:-As we know that, N= 1/ SinC or, SinC=1/N. Or, SinC = 4/3. Or, SinC=3/4

SinC = Sin48.59°. Or, C= 48.59°

So the critical angel is 48.59°.

(5) Determine the refractive index of benzene if the critical angle is 42°.(AS1)

Ans: –Let, critical angel = C =42°

So the refractive index of benzene is = 1/SinC = 1/Sin42° = 1.5

(6) Explain the formation of mirage? (AS1)

Ans: –Mirage is nothing but an optical illusion. The main reason for its formation is the total internal reflection of light. It is generally seen on a humid sunny day where we can see water at a distance but when we go there is nothing.

(7) How do you verify experimentally that sin i /sin r is a constant? (AS1)

Ans: If we draw two perpendicular lines on a paper passing through the centre as shown. Then we will take a protector and draw 0° to 90° along with the line RS on both sides. Now a place a semi-circular glass along the PQ centre as O. Then we pass the light at 20° with PQ and calculate the reflection angel. Repeating this process at different angel we will see the ratio of the sine angel of incidence and reflection will be same. So we came to this conclusion that Sini/ Sinr is correct.

(8) Explain the phenomenon of total internal reflection with one or two activities. (AS1)

Ans: –when a light is passes through the denser to rarer medium, the angle of incident is greater than the critical angle as a result the light rays get reflected instead of refracted. This phenomenon is called the total internal reflection.

Example: –Formation of mirage.

(9) How do you verify experimentally that the angle of refraction is more than angle of incidence when light rays travel from denser to rarer medium. (AS1)

Ans: –Let’s take a glass slab as a denser medium .Then if we pass the light through it we will see that light ray moves away from the normal line and create an angel called refraction angel. If now we measure the angle we will see the angle of reflection is more than the angle of the incident.

(10) Take a bright metal ball and make it black with soot in a candle flame. Immerse it in water.How does it appear and why? (Make hypothesis and do the above experiment). (AS2)

Ans: –A metal ball which coated with candle flame when it immerse in water we see that it’s shining. As layer of ball became a very thin layer if air acts as rarer medium to water (denser). As a result of it total internal reflection takes place and the ball seems shining.

(11) Take a glass vessel and pour some glycerin into it and then pour water up to the brim. Take a quartz glass rod. Keep it in the vessel. Observe the glass rod from the sides of the glass vessel.

• What changes do you notice?
• What could be the reasons for these changes? (AS2)

Ans: –By applying this process we will see that the glass rod size will increase in water but it will be disappeared in glycerin. The reason behind this phenomenon is the refractive index, as both refractive index is close to same. There will be no bending as light speed is same for same refractive index.

(12) Do activity-7 again. How can you find critical angle of water? Explain your steps briefly. (AS3)

Ans:- For finding the critical angel of the in the of water we first put some water in it as without the water the reflection from the coin would not appear to our eyes as it would blocked by the walls of cylinder. After pouring the reflection ray comes to the surface of the water the we could measure the critical angel of the water.

(13) Collect the values of refractive index of the following media. (AS4) Water, coconut oil, flint glass, crown glass, diamond, benzene and hydrogen gas.

Ans: –Refractive index of this medium is water = 1.33; coconut oil= 1.44; flint gas= 1.65; crown glass = 1.52; diamond = 2.42; benzene= 1.5; hydrogen gas = 1.

(15) Take a thin thermocol sheet. Cut it in circular discs of different radii like 2cm, 3cm,4cm, 4.5cm,5cm etc. and mark centre’s with sketch pen. Now take needles of length nearly 6cm. Pin a needle to each disc at its centre vertically. Take water in a large opaque tray and place the disc with 2cm radius in such a way that the needle is inside the water as shown in fig-Q15. Now try to view the free end (head) of the needle from surface of the water.

• Are you able to see the head of the needle? Now do the same with other discs of different radii. Try to see the head of the needle, each time.

Note: the position of your eye and the position of the disc on water surface should not be changed while repeating the activity with other discs.

• At what maximum radius of disc, were you not able to see the free end of the needle?
• Why were you not able to view the head of the nail for certain radii of the discs?
• Does this activity help you to find the critical angle of the medium (water)?
• Draw a diagram to show the passage of light ray from the head of the nail in different situations. (AS4)

Ans: If the angel of incident is less than the critical angel then we will be able to see the Needle. From the question we can measure that the critical angel is 48.59 which is the critical angel of water. So the value is 48.59 then the needle would not be seen.

(16) Explain the refraction of light through a glass slab with a neat ray diagram. (AS5)

Ans: The ray when incident on the surface on the glass it bend away from the normal as we know that light ray passing through rarer to denser medium it bends. After entering to the glass it follow the path OE and passes from denser to rarer or glass to air by EZ way. Here the incident angel is i, reflected angel is r.

(17) Place an object on the table. Look at the object through the transparent glass slab. You will observe that it will appear closer to you. Draw a ray diagram to show the passage of light ray in this situation. (AS5)

Ans: From the diagram we can see that the image appears closer to us by seeing from the glass slab.

(18) What is the reason behind the shining of diamonds and how do you appreciate it? (AS6)

Ans: –The reason behind the shining of the diamonds is the total internal reflection of diamond. As the diamond has high refractive index it doesn’t reflect more rays.For this condition the diamond always shine.

(19) How do you appreciate the role of Fermat principle in drawing ray diagrams? (AS6)

Ans: –Fermat’s principal states that light always need shortest path to travel. It also helps to find the image from by mirror and also explain laws of reflection, retraction.

(20) A light ray is incident on air-liquid interface at 450 and is refracted at 300. What is the refractive index of the liquid? For what angle of incidence will the angle between reflected ray and refractedray be 900? (AS7)

Ans: –Refractive index = Sini/Sinr

Here i= 45° and r= 30° then refractive index = Sin45°/ Sin30 = 1.14

By questions angel of reflection(i) + angel of incident(r) = 90° or i + r= 90° or r= 90° – i

So n= Sini/ Sin(90°- (i) = Sini/ Cosi = tani

Or, tani=1.14 or, i= 54.7°

The incident angel will be 54.7°.

(21) Explain why a test tube immersed at a certain angle in a tumbler of water appears to have a mirror surface for a certain viewing position? (AS7)

Ans: –The light rays in water strike the test tube at an angel greater than critical angel. For this reason total internal reflection takes places. When these rays come to our eyes it seems from the surface of it.

(22) What is the angle of deviation produced by a glass slab? Explain with ray diagram. (AS7)

Ans: – The angel of deviation produced by glass slab is 0° as the emigrant light rays parallel to the incident rays.

(23) In what cases does a light ray not deviate at the interface of two media? (AS7)

Ans: –The does not deviate if refractive index are equal and if the incidence ray strikes at incidence point.

(24) A ray of light travels from an optically denser to rarer medium. The critical angle of the two media is ‘c’. What is the maximum possible deviation of the ray? (AS7)

Ans: –When the critical angel of two media is c then maximum possible deviation will ne π-c.

(25) When we sit at a camp fire, objects beyond the fire are seen swaying. Give the reason for it. (AS7)

Ans: –The denser air causing the refract of the light beyond the fire at a camp fire. For this reason the object light get reflected and we see swaying.

(26) Why do stars appear twinkling? (AS7)

Ans: –There are various particles in different atmosphere layer as the light rays passes through this layer the light rays get reflected and we see stars appear twinkling.

(27) Why does a diamond shine more than a glass piece cut to the same shape? (AS7)

Ans: –The refractive index of diamond is higher than the glass and for which more total internal reflection takes places in diamond. For that reason we see diamond shine more than the glass piece.

FILL. IN THE BLANKS

(1) At critical angle of incidence, the angle of refraction is…..90°………..

(2) n1 sini = n2 sin r, is called……. Snell’s law…….

(3) Speed of light in vacuums is ………..30000Km/s…

(4) Total internal reflection takes place when a light ray propagates from …denser……. to ….rarer……… medium.

(5) The refractive index of a transparent material is 3/2. The speed of the light in that medium is…..2×10^8 m/s….

(6) Mirage is an example of …….total internal reflection……

MULTIPLE CHOICE QUESTIONS

(1) Which of the following is Shell’s law.

->(b) n1Sini = n2Sinr.

(2) The refractive index of glass with respect to air is 2. Then the critical angle of glass-air interface is………

(c) 30°

(3) Total internal reflection takes place when the light ray travels from……..denser to rarer medium…….

(4) The angle of deviation produced by the glass slab is…….

(a) 0°……

Updated: November 12, 2021 — 1:26 pm