Q: Question 13The refractive index of water is 4/3 and that of glass is 3/2. The refractive index of glass with respect to water is:9/828/91

9/8Reason — Given,aμw = 43\dfrac{4}{3}34aμg = 32\dfrac{3}{2}23wμg = Refractive index of glass Refractive index of water\dfrac{\text{Refractive index of glass }}{\text {Refractive index of water}}Refractive index of waterRefractive index of glass Substituting we get,wμg = 3/2 4/3\dfrac{\text{3/2 }}{\text {4/3}}4/33/2 = 3 x 3 2 x 4\dfrac{\text{3 x 3 }}{\text {2 x 4}}2 x 43 x 3 = 98\dfrac{9}{8}89

Q: Question 4A ray of light of wavelength 6600 Å suffers refraction from air to glass. Taking aμg = 3/2, find the wavelength of light in glass.

As we know that,aμg=wavelength of light in airwavelength of light in glass{\text{a}}\text{μ}{\text{g}} = \dfrac{\text {wavelength of light in air}}{\text {wavelength of light in glass}} \\[0.5em]aμg=wavelength of light in glasswavelength of light in airGiven,Refractive index of glass with respect to air is given by aμg = 32\dfrac{3}{2}23Wavelength of light in air = 6600 ÅSubstituting the values in the formula above we get,32=wavelength of light in airwavelength of light in glass32=6600wavelength of light in glasswavelength of light in glass=660032wavelength of light in glass=6600×23wavelength of light in glass=4400\dfrac{3}{2} = \dfrac{\text {wavelength of light in air}}{\text{wavelength of light in glass}} \\[0.5em] \dfrac{3}{2} = \dfrac{6600}{\text {wavelength of light in glass}} \\[0.5em] \text {wavelength of light in glass} = \dfrac{6600}{\dfrac{3}{2}} \\[0.5em] \text {wavelength of light in glass} = 6600 \times \dfrac{2}{3} \\[0.5em] \text {wavelength of light in glass} = 4400 \\[0.5em]23=wavelength of light in glasswavelength of light in air23=wavelength of light in glass6600wavelength of light in glass=236600wavelength of light in glass=6600×32wavelength of light in glass=4400Hence, wavelength of light in glass = 4400 Å

Question 1

Question 7In the adjacent diagram, AO is a ray of light incident on a rectangular glass slab.(a) Complete the path of the ray until it emerges out of the slab.(b) In the diagram, mark the angle of incidence (i) and the angle of refraction (r) at the first interface. How is the refractive index of glass related to angles i and r?(c) Mark angle of emergence by the letter e. How are the angles i and e related?(d) Which two rays are parallel to each other? Name them.(e) Indicate in the diagram the lateral displacement between the emergent ray and the incident ray. State one factor that affects the lateral displacement.

Accepted Answer

(a) The below diagram shows the path of the ray until it emerges out of the slab:(b) In diagram, the angle of incidence (i) and the angle of refraction (r) are marked.Refractive index of glass μ is related to the angle of incidence (i) and angle of reflection (r) by Snell's second law and is represented as —The ratio of the sine of the angle of incidence i to the sine of the angle of refraction r is constant for the pair of given mediasin isin r=constant 1μ2\dfrac{\text{sin i}}{\text{sin r}} = \text{constant } _1\mu_2 \[0.5em]sin rsin i​=constant 1​μ2​The constant 1μ2 is called the refractive index of the second medium with respect to the first medium.(c) The angle of emergence (e) is marked in diagram.As the refraction occurs at two parallel surfaces PQ and RS, therefore angle MOB = angle N1BO and by the principle of reversibility of the path of a light ray angle of emergence (e) is equal to the angle of incidence (i).(d) As the refraction occurs at two parallel surfaces PQ and RS, therefore angle MOB = angle N1BO and by the principle of reversibility of the path of a light ray angle of emergence (e) is equal to the angle of incidence (i).Hence, the emergent ray BC is parallel to the incident ray AO.(e) The lateral displacement between the incident ray and the emergent ray is represented by XY.One of the factors that affects lateral displacement is the the thickness of the medium. More the thickness of the medium more is the lateral displacement.

Question 2

Question 8A ray of green light enters a liquid from air, as shown in the figure. The angle 1 is 45° and angle 2 is 30°.(a) Find the refractive index of liquid.(b) Show in the diagram the path of the ray after it strikes the mirror and re-enters in air. Mark in the diagram the angles wherever necessary.(c) Redraw the diagram if plane mirror becomes normal to the refracted ray inside the liquid. State the principle used.

Accepted Answer

(a) Refractive index of the liquid is given by Snell's law and is shown as below,airμliquid = sin isin r\dfrac{\text{sin i}}{\text{sin r}}sin rsin i​airμliquid = sin 45°sin 30°\dfrac{\text{sin 45}\degree}{\text{sin 30}\degree}sin 30°sin 45°​=1212=22=2=1.414= \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{2}} \[0.5em] = \dfrac{2}{\sqrt{2}} \[0.5em] = \sqrt{2} \[0.5em] = 1.414 \[0.5em]=21​2​1​​=2​2​=2​=1.414(b) Below labelled diagram shows the path of the ray after it strikes the mirror and re-enters in air:(c) Below labelled diagram shows the path of the ray when it strikes the plane mirror normally inside the liquid:The principle of reversibility is used.

Question 3

Question 13The refractive index of water is 4/3 and that of glass is 3/2. The refractive index of glass with respect to water is:9/828/91

Accepted Answer

9/8Reason — Given,aμw = 43\dfrac{4}{3}34​aμg = 32\dfrac{3}{2}23​wμg = Refractive index of glass Refractive index of water\dfrac{	ext{Refractive index of glass }}{	ext {Refractive index of water}}Refractive index of waterRefractive index of glass ​Substituting we get,wμg = 3/2 4/3\dfrac{	ext{3/2 }}{	ext {4/3}}4/33/2 ​ = 3 x 3 2 x 4\dfrac{	ext{3 x 3 }}{	ext {2 x 4}}2 x 43 x 3 ​ = 98\dfrac{9}{8}89​

Question 4

Question 4A ray of light of wavelength 6600 Å suffers refraction from air to glass. Taking aμg = 3/2, find the wavelength of light in glass.

Accepted Answer

As we know that,aμg=wavelength of light in airwavelength of light in glass{\text{a}}\text{μ}{\text{g}} = \dfrac{\text {wavelength of light in air}}{\text {wavelength of light in glass}} \[0.5em]aμg=wavelength of light in glasswavelength of light in airGiven,Refractive index of glass with respect to air is given by aμg = 32\dfrac{3}{2}23Wavelength of light in air = 6600 ÅSubstituting the values in the formula above we get,32=wavelength of light in airwavelength of light in glass32=6600wavelength of light in glasswavelength of light in glass=660032wavelength of light in glass=6600×23wavelength of light in glass=4400\dfrac{3}{2} = \dfrac{\text {wavelength of light in air}}{\text{wavelength of light in glass}} \[0.5em] \dfrac{3}{2} = \dfrac{6600}{\text {wavelength of light in glass}} \[0.5em] \text {wavelength of light in glass} = \dfrac{6600}{\dfrac{3}{2}} \[0.5em] \text {wavelength of light in glass} = 6600 \times \dfrac{2}{3} \[0.5em] \text {wavelength of light in glass} = 4400 \[0.5em]23=wavelength of light in glasswavelength of light in air23=wavelength of light in glass6600wavelength of light in glass=236600wavelength of light in glass=6600×32wavelength of light in glass=4400Hence, wavelength of light in glass = 4400 Å

Question 5

Question 5(a) Compare the speeds of light of wavelength 4000Å (i.e. violet light) and 8000Å (i.e. red light) in vacuum.(b) How is the refractive index of a medium related to the speed of light in it and in vacuum or air?

Accepted Answer

(a) As we know that light is an electromagnetic wave and it travels with same speed for all wavelength, which is equal to 3 x 108.Hence, the ratio of speeds of light of wavelength 4000Å (i.e. violet light) and 8000Å (i.e. red light) in vacuum = 1:1.(b) The refractive index of a medium is defined as the ratio of the speed of light in vacuum (or air) to the speed of light in that medium.μ=Speed of light in vacuum or air (c) Speed of light in that medium (V)μ =\dfrac{\text{Speed of light in vacuum or air (c) }}{\text {Speed of light in that medium (V)}} \[0.5em]μ=Speed of light in that medium (V)Speed of light in vacuum or air (c) ​Denser medium has a higher refractive index. Hence, the speed of light in such medium is lower in comparison to the speed of light in a medium which has a lower refractive index.

Question 6

Question 9A boy uses blue colour of light to find the refractive index of glass. He then repeats the experiment using red colour of light. Will the refractive index be the same or different in the two cases? Give a reason to support your answer.

Accepted Answer

The speed of blue light in glass is less that of red light.i.e. cb < crWe know that,μ=Speed of light in vacuum or air (c) Speed of light in glass (V)μ = \dfrac{\text{Speed of light in vacuum or air (c) }}{\text {Speed of light in glass (V)}} \[0.5em]μ=Speed of light in glass (V)Speed of light in vacuum or air (c) ​Hence, the refractive index of blue light is greater than that of red light.i.e. μb > μr.Hence, the refractive index of red and blue light in glass will be different.

Question 7

Question 10Name two factors on which the refractive index of a medium depends? State how does it depends on the factors stated by you.

Accepted Answer

The factors on which the refractive index of a medium depends are as follows —(i) Nature of the medium (on the basis of speed of light) — Less the speed of light in the medium as compared to that in air, more is the refractive index of the medium.We know that,μ=Speed of light in vacuum or air (c) Speed of light in glass (V)μ = \dfrac{\text{Speed of light in vacuum or air (c) }}{\text {Speed of light in glass (V)}} \[0.5em]μ=Speed of light in glass (V)Speed of light in vacuum or air (c) ​Vglass = 2 x 10 8 ms-1, μglass = 1.5Vwater = 2.25 x 10 8 ms-1, μwater = 1.33(ii) Physical condition such as temperature — With an increase in temperature, the speed of light in medium increases, so the refractive index of the medium decreases.

Question 8

Question 12When an illuminated object is held in front of a thick plane glass mirror, several images are seen, out of which the second image is the brightest. Give reason.

Accepted Answer

As shown in the figure, if an illuminated object is held in front of a thick plane glass mirror, and is viewed obliquely, a number of images are seen and out of these the second image is the brightest.The reason for this is that when a ray of the light falls on the surface of the mirror from the source, a small part of light (nearly 4%) is reflected forming a faint virtual image, while a large part of light (nearly 96%) is refracted inside the glass.Now, we can see that this ray is strongly reflected back by the silvered surface inside the glass. This ray is then partially refracted in air and this refracted ray forms another virtual image.This image is the brightest image because it is due to the light suffering a strong first reflection at the silvered surface.

Question 9

Question 1Prove that,Refractive index=Real depthApparent depth	ext {Refractive index} = \dfrac {	ext {Real depth}}{	ext {Apparent depth}}Refractive index=Apparent depthReal depth​

Accepted Answer

Consider a point object O kept at the bottom of a transparent medium ( such as water or glass ) separated from air by the surface PQ.A ray of light OA, starting from the object O, is incident on the surface PQ normally, so it passes undeviated along the path AA'. Another ray OB, starting along the object O, strikes the boundary surface PQ at B and suffers refraction.Since, the ray travels from denser to rarer medium so it bends away from normal N'BN drawn at the point of incidence B on the surface PQ and travels along BC in air.When viewed by the eye , the ray BC appears to be coming from point I which is the virtual image of O, obtained on producing A'A and CB backwards.Thus, any point seen from air will appear to be at I, which is lesser depth = AI than its actual depth AO.Angle of incidence = ∠OBN'Angle of refraction = ∠CBN.Since, AO and BN’ are parallel and OB is transversal line, so∠AOB = ∠OBN' = iSimilarly, IA’ and BN are parallel and IC is the transversal line, so∠BIA’ = ∠CBN = rIn right-angle triangle BAO,sin i=BAOB\text {sin i} = \dfrac{BA}{OB}sin i=OBBA​and,In right-angle triangle IAB,sin r=BAIB\text {sin r} = \dfrac{BA}{IB}sin r=IBBA​For refraction from medium to air, by Snell’s lawmμa=sin isin r=BAOBBAIB⇒mμa=IBOB_mμ _a = \dfrac{\text {sin i}}{\text {sin r}} = \dfrac {\dfrac {BA}{OB}}{\dfrac {BA}{IB}} \[0.5em] \Rightarrow _mμ _a = \dfrac{IB}{OB} \[0.5em]m​μa​=sin rsin i​=IBBA​OBBA​​⇒m​μa​=OBIB​ Hence, refractive index of medium with respect to air is,aμm=1mμa=OBIB_aμ _m = \dfrac{1}{_mμ_a} = \dfrac{OB}{IB} \[0.5em]a​μm​=m​μa​1​=IBOB​The object is viewed from a point vertically above the object O, since point B is very close to the point A. ∴ IB = IA and OB = OAHence,aμm=1mμa=OAIA⇒Real depthApparent depth_aμ _m = \dfrac{1}{_mμ_a} = \dfrac{OA}{IA} \[0.5em] \Rightarrow \dfrac{\text {Real depth}}{\text {Apparent depth}} \[0.5em]a​μm​=m​μa​1​=IAOA​⇒Apparent depthReal depth​

Question 10

Question 3A postage stamp kept below a rectangular glass slab of refractive index 1.5 when viewed from vertically above it, appears to be raised by 7.0 mm. Calculate the thickness of the glass slab.

Accepted Answer

As we know,Shift=Real depth×(1–1aμm)\text {Shift} = \text {Real depth} \times (1 – \dfrac{1}{_aμ_m})Shift=Real depth×(1–a​μm​1​)Given,Shift in the image = 7 mm or 0.7 cmRefractive index of the glass block = 1.5So, substituting the values in the formula we get,0.7=Real depth×(1–11.5)0.7=Real depth×0.51.5Real depth=0.7×1.50.5⇒Real depth=2.10.7 = \text {Real depth} \times (1 – \dfrac{1}{1.5}) \[0.5em] 0.7 = \text {Real depth} \times \dfrac{0.5}{1.5} \[0.5em] \text {Real depth} = \dfrac{0.7 \times 1.5}{0.5} \[0.5em] \Rightarrow \text {Real depth} = 2.1 \[0.5em]0.7=Real depth×(1–1.51​)0.7=Real depth×1.50.5​Real depth=0.50.7×1.5​⇒Real depth=2.1Hence, the thickness of glass slab = 2.1 cm.(Thickness of glass slab is same as the real depth of the postage stamp).

Question 11

Question 6The refractive index of glass is 1.5. From a point P inside a glass slab, draw rays PA, PB and PC incident on the glass-air surface at an angle of incidence 30°, 42° and 60° respectively.(a) In the diagram show the approximate direction of these rays as they emerge out of the slab.(b) What is the angle of refraction for the ray PB?(Take sin 42° = 2 / 3)

Accepted Answer

(a) Diagram with rays PA, PB and PC is shown below:(b) Given,μg = 1.5As we know,sin ic=1μ\text {sin i}_c = \dfrac {1}{μ} \[0.5em]sin ic​=μ1​Substituting the values in the formula we get,sin ic=11.5⇒sin ic=0.667⇒ic=41.8\text {sin i}_c = \dfrac{1}{1.5} \[0.5em] \Rightarrow \text {sin i}_c = 0.667 \[0.5em] \Rightarrow i_c = 41.8sin ic​=1.51​⇒sin ic​=0.667⇒ic​=41.8Hence, we can round off ic = 42°Applying,sin rsin i=aμgsin r=aμg×sin isin r=aμg×sin 42°\dfrac{\text {sin r}}{\text {sin i}} = _a\mu_g \[0.5em] \text {sin r} = _a\mu_g \times \text {sin i} \[0.5em] \text {sin r} = _a\mu_g × \text {sin 42°} \[0.5em]sin isin r​=a​μg​sin r=a​μg​×sin isin r=a​μg​×sin 42°Given,sin 42°=23\text {sin 42°} = \dfrac{2}{3}sin 42°=32​andaμg=32_aμ_g = \dfrac{3}{2}a​μg​=23​Substituting the values in the formula we get,sin r=32×23⇒sin r=1\text{sin r} = \dfrac{3}{2} \times \dfrac{2}{3} \[0.5em] \Rightarrow \text {sin r} = 1 \[0.5em]sin r=23​×32​⇒sin r=1As sin r = 1 , so angle of refraction = 90°.

Question 12

Question 11Draw a diagram of a right angled isosceles prism which is used to make an inverted image erect.

Accepted Answer

Below diagram shows the usage of a right angled isosceles prism which is used to make an inverted image erect:This action of prism is used in a slide.

Question 13

Question 14Draw a neat labelled ray diagram to show total internal reflection of a ray of light incident normally on one face of a 30°, 90°, 60° prism.

Accepted Answer

Below is the labelled ray diagram showing total internal reflection of a ray of light incident normally on one face of a 30°, 90°, 60° prism

Question 14

Question 15Copy the diagram given below and complete the path of the light ray till it emerges out of the prism. The critical angle of glass is 42°. In your diagram mark the angles wherever necessary.

Accepted Answer

Below is the completed diagram showing the path of the light ray till it emerges out of the prism with all angles marked:

Question 15

Question 16What device other than a plane mirror, can be used to turn a ray of light through 180°? Draw a diagram in support of your answer. Name an instrument in which this device is used.

Accepted Answer

A total reflecting prism is used to turn a ray of light through 180°.This device is used in a binocular and camera to invert the image without the loss of intensity.

Question 16

Question 17Two isosceles right-angled glass prisms P and Q are placed near each other as shown in figure. Complete the path of the light ray entering the prism P till it emerges out of the prism Q.

Accepted Answer

Below diagram shows the path of the light ray entering the prism P till it emerges out of the prism Q:

Question 17

Question 18Complete the path of ray PQ through the glass prism ABC as shown in figure till it emerges out of the prism. Given the critical angle of glass is 42°.

Accepted Answer

Below diagram shows the path of ray PQ through the glass prism ABC till it emerges out of the prism:

Question 18

Question 19The diagram below shows a light source P embedded in a rectangular glass block ABCD of critical angle 42°. Complete the path of ray PQ till it emerges out of the block. [Write necessary angles]

Accepted Answer

Completed diagram showing the path of the ray PQ till it emerges out of the block is shown below:

Question 19

Question 2State the approximate value of the critical angle for(a) glass-air surface(b) water-air surface.

Accepted Answer

(a) As we know,aμg=1sin C=cosec C_aμ_g = \dfrac{1}{\text {sin C}} \[0.5em] = \text {cosec C}a​μg​=sin C1​=cosec CRefractive index isaμg=32_aμ_g = \dfrac{3}{2}a​μg​=23​Substituting the values in the formula we get,32=1sin Csin C=132sin C=23\dfrac{3}{2} = \dfrac{1}{\text {sin C}} \[1em] \text {sin C} = \dfrac{1}{\dfrac{3}{2}} \[1em] \text {sin C} = \dfrac{2}{3}23​=sin C1​sin C=23​1​sin C=32​As,sin 42°=23\text {sin 42°} = \dfrac {2}{3}sin 42°=32​Hence, critical angle for glass air surface is 42°.(b) As we know,aμw=1sin C=cosec C_aμ_w = \dfrac{1}{\text {sin C}} = \text {cosec C} \[0.5em]a​μw​=sin C1​=cosec CRefractive index isaμw=43_aμ_w = \dfrac {4}{3} \[0.5em]a​μw​=34​Substituting the values in the formula we get,43=1sin Csin C=143sin C=34sin 49°=34\dfrac{4}{3} = \dfrac{1}{\text {sin C}} \[1em] \text {sin C} = \dfrac{1}{\dfrac{4}{3}} \[1em] \text {sin C} = \dfrac{3}{4} \[1em] \text {sin 49°} = \dfrac{3}{4} \[0.5em]34​=sin C1​sin C=34​1​sin C=43​sin 49°=43​Hence, critical angle for water air surface is 49°.

Refraction of Light at Plane Surfaces

Exercise 4A Long Questions

Exercise 4A Multiple Choice Type

Exercise 4A Numericals

Exercise 4A Short Questions

Exercise 4C Long Questions

Exercise 4C Numericals

Exercise 4D Long Questions

Exercise 4D Very Short Questions