Q 1 :

A light ray travelling in a glass medium is incident on a glass-air interface at an angle of incidence θ. The reflected (R) and transmitted (T) intensities, both as functions of θ, are plotted. The correct sketch is                     [2011]

  •  

  •  

  •  

  •  

(3)

When the light is incident on a glass–air interface while travelling from glass at an angle less than the critical angle, a small part of the light will be reflected and most part will be transmitted.

When the light is incident at an angle greater than the critical angle, it undergoes complete reflection (total internal reflection), resulting in 0% transmission and 100% reflection.

These characteristics are depicted in option (3).



Q 2 :

A light beam is travelling from Region I to IV (figure). The refractive index in regions I, II, III and IV are n0, n02, n06 and n08 respectively. The angle of incidence θ for which the beam just misses entering region IV is                 [2008]

  • sin-1(34)

     

  • sin-1(18)

     

  • sin-1(14)

     

  • sin-1(13)

     

(2)

For refraction at parallel interfaces,

From Snell's law,

       n0sinθ=n02sinα=n06sinβ=n08sin90°

The angle of refraction in region IV must be 90° as the beam just misses entering region IV.

   sinθ=18

or,  θ=sin-1(18)



Q 3 :

A container is filled with water (μ=1.33) up to a height of 33.25 cm. A concave mirror is placed 15 cm above the water level and the image of an object placed at the bottom is formed 25 cm below the water level. The focal length of the mirror is                 [2005]

  • 15 cm

     

  • 20 cm

     

  • −18.31 cm

     

  • 10 cm

     

(3)

The image I' for first refraction (i.e., when the ray comes out of the liquid) is at a depth of

=33.251.33=25 cm              [ Apparent depth=Real depthμ]

Now, reflection will occur at the concave mirror. For this, I' behaves as an object.

Distance of object from mirror, u=-(15+25)=-40 cm

and  v=-(15+251.33)

where 251.33 is the real depth of the image.

Using mirror formula,

     1f=1v+1u1f=1-33.8+1-40

  f=-18.31 cm



Q 4 :

A point object is placed at the centre of a glass sphere of radius 6 cm and refractive index 1.5. The distance of virtual image from the surface is         [2004]

  • 6 cm

     

  • 4 cm

     

  • 12 cm

     

  • 9 cm

     

(1)

Distance of virtual image from the surface = 6 cm.

The rays coming from the point object fall normally on the glass-air interface and hence pass undeviated. Therefore, if we retrace the path of the refracted rays backwards, the image will be formed at the centre only.



Q 5 :

A source emits sound of frequency 600 Hz inside water. The frequency heard in air will be equal to (velocity of sound in water = 1500 m/s, velocity of sound in air = 300 m/s)       [2004]

  • 3000 Hz

     

  • 120 Hz

     

  • 600 Hz

     

  • 6000 Hz

     

(3)

Frequency does not change with a change of medium.

   Frequency of sound in water = Frequency heard in air = 600 Hz.



Q 6 :

A beam of white light is incident on a glass-air interface from glass to air such that green light just suffers total internal reflection. The colors of the light which will come out to air are                             [2004]

  • Violet, Indigo, Blue

     

  • All colors except green

     

  • Yellow, Orange, Red

     

  • White light

     

(3)

sinθc=1μ  and  μ1λ      sinθcλ

For higher values of λ, the critical angle θc also increases.

Hence, yellow, orange and red colours of light for which θc>incidence angle will come out to air.



Q 7 :

Which one of the following spherical lenses does not exhibit dispersion? The radii of curvature of the surfaces of the lenses are as given in the diagrams.          [2002]

  •  

  •  

  •  

  •  

(3)

Since both surfaces have the same radius of curvature, R1=R2=R on the same side, no dispersion will occur.

For no dispersion, 1f=(μ-1)(1R1-1R2)=0



Q 8 :

Sunlight of intensity 1.3 kWm-2 is incident normally on a thin convex lens of focal length 20 cm. Ignore the energy loss of light due to the lens and assume that the lens aperture size is much smaller than its focal length. The average intensity of light, in kWm-2, at a distance 22 cm from the lens on the other side is _______.            [2018]



(130)

Let D be the initial area covered by light and d be the final area covered by light at 22 cm.

From the figure, AFB and CFD are similar.

  dD=220=110

  Ratio of area=d2D2=1100

As there is no energy loss,   I0D2=Id2

  Average intensity of light at a distance of 22 cm

I=D2d2I0=1.3×100=130.00 kW m-2



Q 9 :

Two identical concave mirrors each of focal length f are facing each other as shown in the schematic diagram. The focal length f is much larger than the size of the mirrors. A glass slab of thickness t and refractive index n0 is kept equidistant from the mirrors and perpendicular to their common principal axis. A monochromatic point light source S is embedded at the center of the slab on the principal axis, as shown in the schematic diagram. For the image to be formed on S itself, which of the following distances between the two mirrors is/are correct?                    [2025]

  • 4f+(1-1n0)t

     

  • 2f+(1-1n0)t

     

  • 4f+(n0-1)t

     

  • 2f+(n0-1)t

     

Select one or more options

(1, 2)

From the figure, S-t2+t2n0=2f

S+t(1n0-1)=4f

S=4f+(1-1n0)t

So, option (1) is correct.

Also, S-t2+t2n0=f

S=2f+(1-1n0)t

Hence, option (2) is correct.



Q 10 :

Three glass cylinders of equal height H = 30 cm and same refractive index n=1.5 are placed on a horizontal surface as shown in figure. Cylinder I has a flat top, cylinder II has a convex top and cylinder III has a concave top. The radii of curvature of the two curved tops are same (R = 3 m). If H1, H2 and H3 are the apparent depths of a point (X) on the bottom of the three cylinders, respectively, the correct statement(s) is/are :                     [2019]

  • 0.8 cm<(H2=H1)<0.9 cm

     

  • H2>H1

     

  • H3>H1

     

  • H2>H3

     

Select one or more options

(2, 4)

As we know,  μ=real depthapparent depth

Case - I        H1=H1.5=301.5=20 cm

Case - II       -n1u+n2v=n2-n1R

  -1.5-30+1-H2=1-1.5-300         H2=20.68 cm

Case - III         -1.5-30+1-H3=1-1.5300

  1-H3=-1600-120

  H3=19.35 cm



Q 11 :

A thin convex lens is made of two materials with refractive indices n1 and n2, as shown in figure. The radius of curvature of the left and right spherical surface are equal. f is the focal length of the lens when n1=n2=n. The focal length is f+Δf when n1=n and n2=n+Δn. Assuming Δn(n-1) and 1<n<2, the correct statement(s) is/are.                       [2019]

  • If Δnn<0, then Δff>0

     

  • The relation between Δff and Δnn remains unchanged if both the convex surfaces are replaced by concave surfaces of the same radius of curvature.

     

  • For n=1.5, Δn=10-3 and f=20 cm, the value of |Δf| will be 0.02 cm (round off to 2nd decimal place).

     

  • |Δff|<|Δnn|

     

Select one or more options

(1, 2, 3)

1f=(n-1)2Rf=R2(n-1)

1f+Δf=(n-1)R+(n+Δn-1)R=2(n-1)+ΔnR

  f+Δf=R2(n-1)+Δn

 f+Δff=R2(n-1)+Δn×2(n-1)R=2(n-1)[2(n-1)+Δn]

  Δff=2n-2-2n+2-Δn[2(n-1)+Δn]

=-Δn2n-2+Δn=-Δn2(n-1)                  (i)

                                                       [ Δn(n-1)]

From equation (i), if Δnn<0, then Δff>0.

Also, |Δff|>|Δnn|

The relation between Δff and Δnn remains unchanged if convex surfaces are replaced by concave surfaces of the same radius of curvature.

For n=1.5,  Δn=10-3,  f=20 cm,

equation (i) gives

Δf20=-10-32(1.5-1)Δf=-0.02 cm

or,   |Δf|=0.02 cm



Q 12 :

Two identical glass rods S1 and S2 (refractive index = 1.5) have one convex end of radius of curvature 10 cm. They are placed with the curved surfaces at a distance d as shown in the figure, with their axes (shown by the dashed line) aligned. When a point source of light P is placed inside rod S1 on its axis at a distance of 50 cm from the curved face, the light rays emanating from it are found to be parallel to the axis inside S2. The distance d is                 [2015]

  • 60 cm

     

  • 70 cm

     

  • 80 cm

     

  • 90 cm

     

(2)

For refraction in S1,

      -n1u+n2v=n2-n1R-1.5-50+1V=1-1.5-10

V=50 cm

Again, for refraction in S2,

     -n1u+n2v=n2-n1R

      -1-(d-50)+1.5=1.5-110

  1d-50=120d-50=20

  d=70 cm



Q 13 :

A ray OP of monochromatic light is incident on the face AB of prism ABCD near vertex B at an incident angle of 60° (see figure). If the refractive index of the material of the prism is 3, which of the following is (are) correct?                              [2010]

  • The ray gets totally internally reflected at face CD

     

  • The ray comes out through face AD

     

  • The angle between the incident ray and the emergent ray is 90°

     

  • The angle between the incident ray and the emergent ray is 120°

     

Select one or more options

(1, 2, 3)

(1)     Applying Snell's law at P,

         n1sini=n2sinr

        n1sin60°=n2sinθ

        sin60°=3sinθ

         θ=30°=r

         In quadrilateral BCQP,

         60°+(90°+30°)+135°+PQC=360°

        PQC=45°i=45°

          The critical angle for prism - air pair of media,

           C=sin-1(13)   which is less than 45°.

           Therefore, total internal reflection takes place at face CD.

(2)      In QDM,  QMD=180°-(45°+75°)=60°

           Therefore, the angle of incidence of ray QM on AD is 30°.

            This angle is less than the critical angle. Hence the ray emerges out of face AD.

(3)       From the figure, the angle between the incident ray OP and the emergent ray MR is 90°.



Q 14 :

Most materials have the refractive index, n>1. So, when a light ray from air enters a naturally occurring material, then by Snell's law, sinθ1sinθ2=n2n1, it is understood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the refractive index of the medium is given by the relation, n=cv=±εrμr, where c is the speed of electromagnetic waves in vacuum, v its speed in the medium, εr and μr are the relative permittivity and permeability of the medium respectively.

In normal materials, both εr and μr are positive; implying positive n for the medium. When both εr and μr are negative, one must choose the negative root of n. Such negative refractive index materials can now be artificially prepared and are called meta-materials. They exhibit significantly different optical behavior, without violating any physical laws. Since n is negative, it results in a change in the direction of propagation of the refracted light. However, similar to normal materials, the frequency of light remains unchanged upon refraction even in meta-materials.                         [2012]

Q.    For light incident from air on a meta-material, the appropriate ray diagram is

  •  

  •  

  •  

  •  

(3)

Let n1= refractive index of air and

n2= refractive index of meta material which is negative.

From Snell's law, n2n1=sinθ1sinθ2

  n2 is negative

  θ2 is also negative. Hence graph (3) is correct.



Q 15 :

Most materials have the refractive index, n>1. So, when a light ray from air enters a naturally occurring material, then by Snell's law, sinθ1sinθ2=n2n1, it is understood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the refractive index of the medium is given by the relation, n=cv=±εrμr, where c is the speed of electromagnetic waves in vacuum, v its speed in the medium, εr and μr are the relative permittivity and permeability of the medium respectively.

In normal materials, both εr and μr are positive; implying positive n for the medium. When both εr and μr are negative, one must choose the negative root of n. Such negative refractive index materials can now be artificially prepared and are called meta-materials. They exhibit significantly different optical behavior, without violating any physical laws. Since n is negative, it results in a change in the direction of propagation of the refracted light. However, similar to normal materials, the frequency of light remains unchanged upon refraction even in meta-materials.                       [2012]

Q.      Choose the correct statement.

  • The speed of light in the meta-material is v=c|n|

     

  • The speed of light in the meta-material is v=c|n|

     

  • The speed of light in the meta-material is v=c

     

  • The wavelength of the light in the meta-material (λm) is given by λm=λair|n|, where λair is wavelength of the light in air.

     

(2)

Speed of light in a medium, V=Cn

n for a meta-material =|n|

 Speed of light in the meta-material, V=C|n|