Q: An object of height 1.5 mm is kept between two coaxially mounted spherical mirrors one concave and the other convex, both of focal length 15 cm and separated by 40 cm. The object is at 20 cm from the pole of the concave mirror. Considering the first reflection at the concave mirror and a second reflection at the convex mirror, find the position, nature and size of the final image formed.

For the first reflection u=−20 cm,f=−15 cmv=u−fuf=−60 cm m1=−uv=−2060=−3 (image is real inverted and of size 4.5 mm ) For the second reflection u=(60−40)=20 cm(∵ the image due to the first mirror is formed at a distance of 20 cm from the pole of the second mirror in the direction of the incident ray) f=15 cm(∵ convex mirror ) v=u−fuf=60 cm m2=−2060=−3; total magnification is m=m1 m2=9; image is virtual, erect and of size 13.5 mm )

Q: The absolute refractive index of a medium is 2. What is the speed of light inside that medium?

n=vc→v=nc=23×108(∵c=3×108 m s−1)=1.5×108 m s−1 speed of light in the medium =1.5×108 m s−1

Q: The speed of light in water is 2.25×108 m s−1 and in glass is 2×108 m s−1. Calculate the a. Absolute refractive index of water.b. Absolute refractive index of glass.c. The refractive index of glass with respect to water.d. The refractive index of water with respect to glass.

Speed in water vw=2.25×108 m s−1 and speed in glass vg=2×108 m s−1(a) nw=vwc=2.25×1083×108=1.33∴ Absolute refractive index of water is 1.33(b) ng=vgc=2×1083×108=1.5∴ Absolute refractive index of glass is 1.5(c) ng=nwng=(4/3)(3/2)=89(>1)Hence glass is optically denser than water.(d) 8nw=nng1=(9/8)1=98(<1)Hence water is optically rarer than glass.

Q: Calculate the time taken by light to pass normally through a glass plate of thickness 4 mm, given refractive index of glass =1.5

Time t= speed (v)distance(d) d=4 mm=4×10−3 m Speed v=nc=1.53×108=2×108 m s−1 Time taken t=2×1084×10−3=2×10−11 s

Q: A light wave of frequency 5×1014 Hz passes normally through water of refractive index 34. Calculate the number of light waves contained in 3 m depth of water.

f=5×1014 Hz;n=34 The number of wavelengths in 1 m of the medium is called the wave number (vˉ). vˉ=λm1 is the number of waves contained in unit length, where λm= wavelength of light in medium. ∴ No. of waves contained in 3 m depth =3v =3×λm1=λm3 Wavelength in vacuum λ=fc=5×10143×108=0.6×10−6 m ∴ Wavelength in medium λm=nλ=(4/3)0.6×10−6=0.45×10−6=4500A˚ ∴ No. of waves in 3 m depth =λm3=0.45×10−63=6.67×106

Q: For a given pair of media, when angle of incidence is 40∘ then angle of refraction is 25∘ Find the value of angle of refraction when angle of incidence becomes 50∘ in the same medium. Also calculate deviation produced in the two cases.

sinr1sini1=sinr2sini2⇒ sin25∘sin45∘=sinr2sin50∘⇒ sinr2r2 d(I case )d( II case )=sin40∘sin25∘×sin50∘=0.5036=sin−1(0.5036)=30∘141=i−r=40∘−25∘=15∘=i−r=50∘−30∘141=19∘461

Q: For a ray entering glass from vacuum the angle of incidence is 40∘ and the angle of refraction is 25∘. Find the refractive index of glass.

ution n=sinrsini=sin25∘sin40∘=0.42260.6428=1.52

Q: A ray gets refracted from a transparent medium to vacuum as shown in the figure. Find the refractive index of the medium.

To calculate refractive index of the medium, one has to consider the refraction from vacuum to medium. For this imagine the path of the ray reversed i.e., i=35∘ and r=25∘ n=sinrsini=sin25∘sin35∘=1.36

Q: A ray of light enters glass from vacuum. If the angle of incidence is 45∘ find the angle of refraction. (Refractive index of glass 1.5 )

n∴r=sinrsini⇒sinr=nsini=1.5sin45∘=0.4714=sin−1(0.4714)=28∘7

Question 1

An object of height 1.5 mm is kept between two coaxially mounted spherical mirrors one concave and the other convex, both of focal length 15 cm and separated by 40 cm. The object is at 20 cm from the pole of the concave mirror. Considering the first reflection at the concave mirror and a second reflection at the convex mirror, find the position, nature and size of the final image formed.

Accepted Answer

For the first reflection
u=−20 cm,f=−15 cmv=u−fuf​=−60 cm m1​=−uv​=−2060​=−3​
(image is real inverted and of size 4.5 mm )
For the second reflection
u=(60−40)=20 cm(∵ the image due to the first mirror is formed at a distance of 20 cm from the pole of the second mirror in the direction of the incident ray)
f=15 cm(∵ convex mirror )
v=u−fuf​=60 cm
m2​=−2060​=−3; total magnification is m=m1​ m2​=9;
image is virtual, erect and of size 13.5 mm )

Question 2

The absolute refractive index of a medium is 2. What is the speed of light inside that medium?

Accepted Answer

n​=vc​→v=nc​=23×108​(∵c=3×108 m s−1)=1.5×108 m s−1​
speed of light in the medium =1.5×108 m s−1

Question 3

Light of wavelength 7000A˚ in vacuum is passed through a medium of refractive index 1.5.
(a) What is the frequency of the light in the medium?
(b) What is the wavelength of light in the medium?

Accepted Answer

(a) Frequency in medium is the same as frequency in vacuum as frequency does not change with medium.
∴f​=λc​=7000×10−103×108​ Hz=4.29×1014 Hz​
(b)
n​=λm​λ​→λm​=nλ​=1.57000×10−10​=4667A˚​
Hence the wavelength of light in the medium is 4667A˚.

Question 4

The speed of light in water is 2.25×108 m s−1 and in glass is 2×108 m s−1. Calculate the a. Absolute refractive index of water.b. Absolute refractive index of glass.c. The refractive index of glass with respect to water.d. The refractive index of water with respect to glass.

Accepted Answer

Speed in water vw​=2.25×108 m s−1 and speed in glass vg​=2×108 m s−1(a) nw​=vw​c​=2.25×1083×108​=1.33∴ Absolute refractive index of water is 1.33(b) ng​=vg​c​=2×1083×108​=1.5∴ Absolute refractive index of glass is 1.5(c) ng​=nw​ng​​=(4/3)(3/2)​=89​(>1)Hence glass is optically denser than water.(d) 8​nw​=nng​1​=(9/8)1​=98​(<1)Hence water is optically rarer than glass.

Question 5

Calculate the time taken by light to pass normally through a glass plate of thickness 4 mm, given refractive index of glass =1.5

Accepted Answer

Time t= speed (v)distance(d)​
d=4 mm=4×10−3 m
Speed v=nc​=1.53×108​=2×108 m s−1
Time taken t=2×1084×10−3​=2×10−11 s

Question 6

A light wave of frequency 5×1014 Hz passes normally through water of refractive index 34​. Calculate the number of light waves contained in 3 m depth of water.

Accepted Answer

f=5×1014 Hz;n=34​
The number of wavelengths in 1 m of the medium is called the wave number (vˉ).
vˉ=λm​1​ is the number of waves contained in unit length, where λm​= wavelength of light in medium.
∴ No. of waves contained in 3 m depth =3v
=3×λm​1​=λm​3​
Wavelength in vacuum
λ=fc​=5×10143×108​=0.6×10−6 m
∴ Wavelength in medium
λm​​=nλ​=(4/3)0.6×10−6​=0.45×10−6=4500A˚​
∴ No. of waves in 3 m depth
=λm​3​=0.45×10−63​=6.67×106​

Question 7

For a given pair of media, when angle of incidence is 40∘ then angle of refraction is 25∘ Find the value of angle of refraction when angle of incidence becomes 50∘ in the same medium. Also calculate deviation produced in the two cases.

Accepted Answer

sinr1​sini1​​=sinr2​sini2​​⇒
sin25∘sin45∘​=sinr2​sin50∘​⇒
sinr2​r2​ d(I case )d( II case )​=sin40∘sin25∘×sin50∘​=0.5036=sin−1(0.5036)=30∘141=i−r=40∘−25∘=15∘=i−r=50∘−30∘141=19∘461​

Question 8

For a ray entering glass from vacuum the angle of incidence is 40∘ and the angle of refraction is 25∘. Find the refractive index of glass.

Accepted Answer

ution
n=sinrsini​=sin25∘sin40∘​=0.42260.6428​=1.52

Question 9

A ray gets refracted from a transparent medium to vacuum as shown in the figure. Find the refractive index of the medium.

Accepted Answer

To calculate refractive index of the medium, one has to consider the refraction from vacuum to medium. For this imagine the path of the ray reversed i.e., i=35∘ and r=25∘
n=sinrsini​=sin25∘sin35∘​=1.36

Question 10

A ray of light enters glass from vacuum. If the angle of incidence is 45∘ find the angle of refraction. (Refractive index of glass 1.5 )

Accepted Answer

n∴r​=sinrsini​⇒sinr=nsini​=1.5sin45∘​=0.4714=sin−1(0.4714)=28∘7​

Question 11

Using the data given in the figure, calculate refractive index of glass with respect to water.

Accepted Answer

ng​​=sinrsini​=sin26∘sin30∘​=1.14​
The absolute refractive index of some mediums are as given below.

Question 12

A ray of light falls normal to one face of a right angled isosceles prism and emerges normal to the other face. What is the minimum index of refraction of the glass?

Accepted Answer

45∘>θc​​⇒sin45∘>sinθc​=n1​⇒n>sin451​=1.414​

Question 13

What happens if the above prism (n=1.50) is immersed in water (n=4/3) ?

Accepted Answer

θcglass/water ​=sin−13/2×3/41​=62.7∘>45∘

Hence no total internal reflection, but only refraction takes place
1.5sin45=34​sinr⇒r=52.7∘

Question 14

For a prism, refracting angle is 65∘ and angle of minimum deviation is 40∘. Find
(i) Refractive index of the material of the prism.
(ii) Angle of incidence and angle of emergence corresponding to minimum deviation.
(iii) Angle of refraction corresponding to minimum deviation.

Accepted Answer

(i) n=Sin2A​Sin2A+D​​=Sin265∘​Sin265∘+40​​=Sin32.5∘Sin52.5∘​=1.48
(ii) i
i1​i2​​=2D+A​=240+65​=52.5∘=i1​=52.5∘​
(iii) r1​=2A​
=32.5∘
Note:
When there is minimum deviation, the deviation produced by the first face d1​=(i1​−r1​) is the same as that produced by the second face d2​=(i2​−r2​)

Question 15

A prism of angle 60∘ is of a material of refractive index 2​. Calculate angles of minimum deviation and maximum deviation.

Accepted Answer

(i) A=60∘,n=2​
 At δmin​,r=2A​⇒r=30∘n=sinrsini​⇒2​=sin30∘sini​⇒sini=2​1​⇒i=45∘δmin​=2i−A=90∘−60∘=30∘θc​=sin−12​1​=45∘⇒A>θc​​
(ii) Now, since r1​=θc​ and A=r1​+r2​,
We have r2​=A−r1​=A−θc​
∴i2​=nsinr2​=nsin(A−θc​)
∴ Applicable formula is
δmax​==​90∘+sin−1[nsin(A−θc​)]−A90∘+sin−1[2​.Sin(60∘−45∘)]−60∘=90∘+22∘−60∘=52∘​
(We may also calculate δmax​ without the use of this formula:
 For ∴⇒r2​∴⇒sin2​∴δmax​​δmax​,i1​r1​=θc​=A−r1​=60∘−45∘=15∘sin15∘sini2​​=2​sin15∘=0.37⇒i2​=22∘=i1​+i2​−A=90∘+22∘−60∘=52∘)​=90∘=45∘=2​​

Question 16

If a prism (n=2​) has D=2 A, what is A ?

Accepted Answer

2​=sin2A​sin2A+D​​=sin2A​sin23 A​​
let θ=2A​
⇒2​=sinθsin3θ​=3−4sin2θ
⇒θ=39∘4′⇒A=2θ=78∘8′

Question 17

In one dispersion experiment
dv​=3.25∘,db​=3∘ and
dr​=2.75∘. Calculate angular dispersion, mean deviation and dispersive power for
(i) violet - red pair and (ii) blue - red pair

Accepted Answer

Case I (violet - red pair)
Angular dispersion =dv−dr=3.25∘−2.75∘=0.5∘
Mean deviation =d=2dv​+dr​​=23.25∘+2.75∘​=3∘
Dispersive power =ddv​−dr​​
=3∘0.5∘​=0.167
Case II ( blue - red pair)
Angular dispersion =db​−dr​=3∘−2.75∘ =0.25∘
Mean deviation =d=2db​+dr​​=23∘+2.75∘​=2.875∘
Dispersive power =ddb​−dr​​=2.875∘0.25∘​=0.087

Question 18

Calculate the dispersive power of the material of a prism with respect to the given pair of colours of light. nv​=1.538 and nr​=1.527

Accepted Answer

Mean refractive index n=2nv​+nr​​
=21.538+1.527​=1.5325
Dispersive power ω=n−1nv​−nr​​
=1.5325−11.538−1.527​=0.021

Question 19

A crown glass prism of 11∘ is combined with a flint glass prism to form a combination giving dispersion without deviation. Given the refractive indices as

Find
(1) The angle of the flint glass prism
(2) The net angular dispersion produced

Accepted Answer

Refractive index of mean ray (crown)
n=21.51+1.52​=1.515
Refractive index of mean ray(flint)
n′=21.62+1.64​=1.63
Condition for no deviation is A′A​=−n−1n′−1​
⇒A′​=−n′−1(n−1)A​=−(1.63−1)(1.515−1)11​=−8.99∘​
Angle of flint glass prism =8.99∘

Question 20

An object is immersed in a medium of refractive index n= 2.0. Object is 15 cm from the concave surface whose radius of curvature is 10 cm. Locate the image.

Accepted Answer

u=−15,n1​=2,n2​=1,R=−10
vn2​​−un1​​=Rn2​−n1​​
v1​−−152​=−101−2​⇒v=−30 cm (virtual)

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Solutions for IIT JEE Super Course in Physics Optics and Modern Physics by Trishna Knowledge Systems

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