IIT JEE Physics : Optics previous 15 years questions & answers.

We present here the previous 15 years IIT JEE questions and answers in Optics.   [  pre – draft ]

IIT JEE Advanced & Mains Physics 2018 : Optics : Rank file / expected questions / Physicsmynd Elite Series       

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1. A parallel beam of light is incident from air at an angle \alpha on the side PQ of a right angled triangular prism of refractive index n=\sqrt{{2}}. Light undergoes total internal reflection in the prism at the face PR when \alpha has a minimum value of {45}^{\circ}. The angle \theta of the prism is      (JEE Advanced 2016)

(a) {15}^{\circ}
(b) {30}^{\circ}
(c) {22.5}^{\circ}
(d) {45}^{\circ}

Solution
(a) The refraction of the incident beam at the side PQ of the given right-angled prism POR is
{\frac}{{\sin{\alpha }}}{{\sin{r}}}=n
\Rightarrow{\sin{45^{\circ}}}{{\sin{r}}}=\sqrt{{2}}
\Rightarrow{\sin{r}}=\frac{1}{2} or  r={30}^{\circ}

Note that the total internal reflection at the side PR of the given prism is
{\sin{C}}=\frac{1}{n}{{\sin}^{{-1}}{\left({\frac {1}{\sqrt{{2}}}}\right)}}=45^{\circ}
where C={45}^{\circ} is the critical angle.
Also , in \Delta{PXY} we have
<P<X<Y={180}^{\circ}
\Rightarrow\theta+({90}^{\circ}+r)+({90}^{\circ}-C)={180}^{\circ}
\Rightarrow\theta=C-r={45}^{\circ}-{30}^{\circ}={15}^{\circ}

 

2. A transparent slab of thickness d has a refractive index n(z) that increases with z. Here, z is the vertical distance inside the slab, measured from the top. The slab is placed between two media with uniform refractive indices {n}_{1} and {n}_{2}(>{n}_{1})  as shown in the figure. A ray of light is incident with angle {\theta}_{i} from medium 1 and emerges in medium 2 with refraction angle {\theta}_{f} with a lateral displacement l. Which of the following statement(s) is (are) true?    (JEE Advanced 2016)

(a) l is independent of {n}_{2}
(b) l is dependent on n(z)
(c) {n}_{1}{\sin{{\theta }_{{i}}}}=({n}_{2}-{n}_{1}){\sin{{\theta }_{{i}}}}
(d) {n}_{1}{\sin{{\theta }_{{i}}}}={n}_{2}{\sin{{\theta }_{{f}}}}

Solution
(a), (b), (d) From Snell’s law, we get
{n}_{1}{\sin{{\theta }_{{i}}}}=n(z){\sin{\left[{\theta \left({z}\right)}\right]}}={n}_{2}{\sin{{\theta }_{{f}}}}
In case , there was no slab, the incident ray will go straight and the lateral displacement at distance d would be
l=d{\tan{{\theta }_{{i}}}}
But ,due to  refraction, the refracted ray changes direction, thus changing the lateral displacement l. Since l is measured before the start of medium 2, there is no effect of {n}_{2} on l. Hence, l is independent of {n}_{2} but it is dependent of n(z).
Hence , options (A), (B) and (D) are the correct choices.

 

3. A plano-convex lens is made of a material of refractive index n. When a small object is placed 30 cm away in front of the curved surface of the lens, an image of double the size of the object is produced. Due to reflection from the convex surface of the lens, another faint image is observed at a a distance of 10cm away from the lens. Which of the following statement(s) is(are) true?    (JEE Advanced 2016)

(a) The refractive index of the lens is 2.5.
(b) The radius of curvature of the convex surface is 45cm.
(c) The faint image is crect and real.
(d) The focal length of the lens is 20cm.

Solution
(a), (d)  .  Two cases are to be considered –
Case I:

Magnification, M=\frac{-v}{u}
\Rightarrow{v}=-2\times(-30)
v=60 cm         (1)
On using lens formula, we get
\frac{1}{{f}_{L}}=\frac{1}{60}-\frac{1}{-30}=\frac{1}{20}
\Rightarrow{f}_{L}=20 cm.
Case II: (Convex mirror)

 

We get virtual, erect and diminished image:
\frac{1}{{f}_{m}}=\frac{1}{v}+\frac{1}{u}=\frac{1}{10}+\frac{1}{-30}=\frac{1}{15}
\Rightarrow{f}_{m}=15 cm
That is, the radius of curvaature of the convex surface is
R=2{f}_{m}=30 cm
Now, \frac{1}{{f}_{l}}=\frac{(n-1)}{R}
\Rightarrow{n-1}=\frac{3}{2}\Rightarrow{n}\frac{5}{2}=2.5
Therefore , options (a) and (d) are the correct choices.

 

4. A small object is placed 50 cm to the left of thin convex lens of focal length 30 cm. A convex spherical mirror of radius of curvature 100 cm is placed to the right of the lens at a distance of 50cm. The mirror is tilted such that the axis of the mirror is at an angle \theta={30}^{\circ} to the axis of the lens, as shown in the figure. If the origin of the coordinate system is taken to be at the centre of the lens, the coordinates (in cm) of the point (x,y) at which the image id formed are     (JEE Advanced 2016)

(a) (25,{25}\sqrt{{3}})
(b) (50-{25}\sqrt{{3}},25)
(c) (\frac {125}{3},\frac {25}{\sqrt{{3}}})]
(d) (0,0)

Solution
(a) Applying the lens formula:
\frac{1}{f}=\frac{1}{v}-\frac{1}{u}
\Rightarrow\frac{1}{30}=\frac{1}{v}-\frac{1}{-50}\Rightarrow\frac{1}{v}=\frac{1}{30}-\frac{1}{50}=\frac{1}{75}\Rightarrow{v}=75 cm

For convex mirror before tilting:

Now , using the  mirror formula:
\frac{1}{{f}_{m}}=\frac{1}{v}+\frac{1}{u}

where {f}_{m}=\frac{R}{2}=50 cm
\Rightarrow\frac{1}{50}=\frac{1}{v}+\frac{1}{25}\Rightarrow\frac{1}{v}=\frac{-1}{50}
{v}=-50 cm

where {l}_{2} is formed in front of the convex mirror when the convex mirror is titled:


Rotating a mirror by an angle \theta rotates the reflected ray by an anagle 2\theta
In our case the convex mirror is rotated by {30}^{\circ} therefore, the image gets rotated by
{2}\times{30}^{\circ}={60}^{\circ}

Therefore, the coordinates of the point (x,y) at which the image is formed ({I}_{3}) is
[(-50{\cos{60^{\circ}}}+50),(\left{50{\sin{60^{\circ}}}}\right)]=(+25,{25}\sqrt{{3}})

 

5. Consider a concave mirror and a convex lens (refractive index= 1.5) of focal length 10 cm each separated by a distance of 50 cm in air (refractive index = 1) as shown in the figure. An object is placed at a distance of 15 cm from the mirror. Its erect image formed by this combination has magnification {M}_{1}. When the set-up is kept in medium of refractive index {7}{6} the magnification becomes {M}_{2}. The magnitude \mid{\frac {{M}_{{2}}}{{M}_{{1}}}}\mid is————   (JEE Advanced 2015)

Solution
For concave mirror, mirror formula gives
\frac{1}{{v}_{1}}+\frac{1}{{u}_{1}}=\frac{1}{{f}_{1}}
\frac{1}{{v}_{1}}-\frac{1}{15}=\frac{1}{-10}

Therefore,
\frac{1}{v}_{1}=-\frac{1}{30}\Rightarrow{v}_{1}=-30 cm
The magnification is
{M}_{1}=-\frac{{v}_{1}}{{u}_{1}}=-\left({\frac {-30}{-15}}\right)=-2

For convex lens, lens formula gives
\frac{1}{{v}_{2}}-\frac{1}{{u}_{2}}=\frac{1}{{f}_{2}}
\frac{1}{{v}_{2}}-\frac{1}{-20}=\frac{1}{10}
\frac{1}{{v}_{2}}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}
\Rightarrow{v}_{2}=20 cm

The magnification is
{M}_{2}=\frac{{v}_{2}}{{v}_{2}}=\frac{20}{-20}=-1.
Total magnification is
M+{M}_{1}{M}_{2}=(-2)(-1)=2
On increasing the entire set up in the liquid, f of the concave mirror remains unchanged while that of convex lens becomes
{f}_{1}={f}_{2}\frac {\mu -1}{\left({\frac {\mu }{{\mu }_{{1}}}}\right)-1}

where {f}_{2} is the focal length of the lens in air and {f}_{1} is the focal length of the lens in liquid. Now,
\mu=1.5 and {\mu}_{1}=\frac{7}{6}
Therefore,
{f}_{1}=10\times{1.5-1}{\left[{\frac {1\cdot 5}{{\left({\frac {7}{6}}\right)}}\right]-1}}=\frac{5}{\left({\frac {9}{7}}\right)-1}=\frac{35}{2}
Hence,
\frac{1}{{v}_{3}}=\frac{1}{-20}=\frac{2}{35}
\frac{1}{{v}_{3}}=\frac{2}{35}-\frac{1}{20}=\frac{8-7}{140}
The magnification is
{M}_{3}=\frac {{v}_{{3}}}{{\nu }_{{2}}}\frac{1}{140}
The total magnification here is
{M}{'}={M}_{1}{M}_{3}=(-2)(-7)=14
The ratio is
\mid{\frac {{M}^{{′}}}{M}}\mid=\frac {14}{2}=7

 

 

6. Two identical glass rods {S}_{1} and {S}_{2} (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 {S}_{1} 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 indside {S}_{2}. The distance d is    (JEE Advanced 2015)

(a) 60 cm
(b) 70 cm
(c) 80 cm
(d) 90 cm

Solution
(b)

Using Gaussian formula  –
\frac {{\mu }_{{2}}}{v}-\frac {{\mu }_{{1}}}{v}=\frac {{\mu }_{{2}}-{\mu }_{{1}}}{R}
Considering the first rod ,
\frac{1}{{v}_{1}}-\frac{1.5}{-50}=\frac{1.5-1}{10}
Solving,we get {v}_{1}=50 cm. For the second rod, we get
\frac{1.5}{\infty}-\frac{1}{-(d-50)}=\frac{1.5-1}{10}
That is, {v}_{2}\rightarrow{\infty} as rays in the second rod are parallel to the axis. Solving ,  gives d = 70 cm.

 

7. A monochromatic beam of light is incident at {60}^{\circ} on one face of an equilateral prism of refractive index n and emerges from the opposite face making an angle \theta(n) with the normal (see the figure). For n=\sqrt{{3}} the value of \theta is [60]^{\circ} and \frac{d\theta}{dn}. The value of m is ———-.    ( JEE Advanced 2015)

Solution
(2) Using Snell’s law,
{\sin{60^{\circ}}}=n{\sin{{r}_{{1}}}}            ……..(1)
{\sin{{r}_{{1}}}}=\frac {\sqrt{{3}}}{2\times \sqrt{{3}}}=\frac {1}{2}
{r}_{1}={30}^{\circ}

Now ,
n{\sin{{r}_{{2}}}}=1{\sin{\theta }}
and {r}_{1}+{r}_{2}=A={60}^{\circ}
Therefore,
n{\sin{\left({60^{\circ}-{r}_{{1}}}\right)}}=1{\sin{\theta }}             ……..(2)
Differentiating on both sides ,
{\sin{\left({60^{\circ}-{r}_{{1}}}\right)}}-n{\cos{\left({60^{\circ}-{r}_{{1}}}\right)}}\frac {d{r}_{{1}}}{dn}={\cos{\theta \frac {d\theta }{dn}}}
Differnetiating Eq. (1) on both sides, we get
0={\sin{{r}_{{1}}}}+n{\cos{{r}_{{1}}}}\left({\frac {d{r}_{{1}}}{dn}}\right)
0=\frac{1}{2}+\sqrt{{3}}\left({\frac {\sqrt{{3}}}{2}}\right)\left({\frac {d{r}_{{1}}}{dn}}\right)
Therefore,
\frac {d{r}_{{1}}}{dn}=\frac {-1}{3}
Hence, substituting {r}_{1}={30}^{\circ}, we get
\frac {d{r}_{{1}}}{dn}=\frac {-1}{3}
{\sin{30^{\circ}}}-\sqrt{{3}}{\cos{30^{\circ}}}\left({-\frac {1}{3}}\right)={\cos{60^{\circ}}}\left({\frac {d\theta }{dn}}\right)
\Rightarrow\frac{1}{2}\frac {3}{2\times 3}=\frac {1}{2}\left({\frac {d\theta }{dn}}\right)
\Rightarrow\frac {d\theta }{dn}=2

Paragraph for Questions 8 and 9 :

Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index {n}_{1} surrounded by a medium of lower refractive index  {n}_{2}. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media  {n}_{1} and  {n}_{2} as shown in the below figure. All rays with the angle of incidence i less than a particular value  {i}_{m} are confined in the medium of refractive index  {n}_{1}. The numerical aperture (NA) of the structure is defined as {\sin{{\dot{i}}_{{m}}}}

8. For two structures, namely, {S}_{1} with {n}_{1}=\frac {\sqrt{{45}}}{4} and {n}_{2}=\frac{3}{2} and {S}_{2} with {n}_{1}=\frac{8}{5} and {n}_{2}=\frac{7}{5} and takinh the refractive index of water to be \frac{4}{3} and that of air to be 1, the correct option(s) is (are)    (JEE Advanced 2015)

(a) NA of {S}_{1} immersed in water is the same as that of {S}_{2}  immersed in a liquid of refractive index \frac{16}3\sqrt{{15}}
(b) Na of {S}_{1} immersed in  liquid of refractive index \frac {6}{\sqrt{{15}}} is the same as that of {S}_{2} immersed in water.
(c) NA of {S}_{1} placed in air is the same as that of {S}_{2} immersed in liquid of refractive index \frac {4}{\sqrt{{15}}}
(d) Na of {S}_{1} placed in air is the same as that of {S}_{2} placed in water.

Solution
(a), (c)

Note that the total internal reflection is at the core cladding interface if \theta \geq {i}_{{C}} where {i}_{C} is the critical angle. Also,
\frac {\sin{i }}{\sin{r}}=\frac {{n}_{{i}}}{{n}_{{m}}}   (Snell’s law)
(90-r)\geq{i}_{C}
{\sin{\left({90-r}\right)}}\geq {\sin{{i}_{{C}}}}
{\cos{r}}\geq {\sin{{i}_{{C}}}}=\frac {{n}_{{2}}}{{n}_{{1}}}

Now ,as i increases r also increases. Hence \theta=(90-r) decreases. Thus, there is a maximum value of i (say {i}_{m}) beyond which no total internal reflection occurs at the
interface. Here {i}_{m} is the maximum angle of acceptance. Now,
{n}_{{1}}{\sin{{r}_{{m}}}}={n}_{1}{\sin{\left({90-{i}_{{C}}}\right)}}={n}_{1}{\cos{{i}_{{C}}}}
={n}_{1}\sqrt{{1-{{\sin}^{{2}}{{i}_{{C}}}}}}
={n}_{1}\sqrt{{1-\frac {{n}_{{2}}^{{2}}}{{n}_{{1}}^{{2}}}}}
=\sqrt{{{n}_{{1}}^{{2}}-{n}_{{2}}^{{2}}}}
(NA)^{2}={{\sin}^{{2}}{{i}_{{m}}}}=\frac {{n}_{{1}}^{{2}}-{n}_{{2}}^{{2}}}{{n}_{{m}}^{{2}}}
If {n}_{1}=\frac {\sqrt{{45}}}{4}{n}_{2}=\frac{3}{2}{n}_{m}=\frac{4}{3}, we get
NA={\sin{{i}_{{m}}}}=\frac{9}{16}
If {n}_{1}=\frac{8}{5},  {n}_{2}=\frac{7}{5}{n}_{m}=\frac{16}{3\sqrt{{15}}}, we get
NA={\sin{{i}_{{m}}}}=\frac{9}{16}
If {n}_{1}=\frac {\sqrt{{45}}}{4}{n}_{2}=\frac{3}{2}{n}_{m}=1, we have
NA={\sin{{i}_{{m}}}}=\frac{3}{4}
If {n}_{1}=\frac{8}{5},  {n}_{2}=\frac{7}{5}{n}_{m}=\frac{4}{\sqrt{{15}}}, we have
NA={\sin{{i}_{{m}}}}=\frac{3}{4}

9. If two structures of same cross-sectional area, but different numerical apertures {N}{A}_{1} and {N}{A}_{2} ({N}{A}_{2}<{N}{A}_{1}[/latex]) are joined longitudinally, the numerical aperture of the combined structure is

(a) \frac {N{A}_{{1}}N{A}_{{2}}}{N{A}_{{1}}+N{A}_{{2}}}
(b) {N{A}_{{1}}+N{A}_{{2}}}
(c) N{A}_{{1}}
(d) N{A}_{{2}}

Solution

(d) In order for the total internal reflection to happen in both structures, NA should be the least of the two forming the combined structure. Consequently ,Hence, it will be N{A}_{{2}}

 

Fill in the blanks:

1.A slit of width d is placed in front of a lens of focal length 0.5m and is illuminated normally with light of wavelength 5.89 x 10-7 m.  The first diffraction maximum are separated by 2 x 10-3 m.  The width d of the slit is ……..m. ( IIT JEE 1997 )

2. A light of wavelength 6000 Å in air, enters a medium with refractive index 1.5.  Inside the medium is frequency is ……………Hz and its wavelength is ……….Å. ( IIT JEE 2007 )

3.Two thin lenses, when in contact, produce a combination of power + 10 diopters. When they are 0.25 m apart, the power reduces to + 6 diopters.  The focal length of the lenses are ….. m and …..m. ( IIT JEE 1997 )

4.A ray of light is incident normally on one of the faces of a prism of apex angle 30o and refractive index Ö2.  the angle of deviation of the ray is …… degrees. ( IIT JEE 2007 )

  Objective Questions:

(Only one option is correct)

1. Spherical aberration in a thin lens can be reduced by:

  1. Using a monochromatic light
  2. Using a doublet combination
  3. Using a circular annular mark over the lens
  4. Increasing the size of the lens                                 ( IIT JEE 1994 )

2 . A narrow slit of width 1 nm is illuminated by monochromatic light of wavelength 600 nm.  The distance between the first 5t minima on either side of a screen at a distance of 2 m is:   ( IIT JEE 1994 )

  1. 1.2 cm
  2. 1.2 mm
  3. 2.4 cm
  4. 2.4 mm

3.An isosceles prism of angle 120o has a refractive index 1.44. Two parallel beams of monochromatic light enter the prism parallel to each other in air   .  The rays emerge from the opposite face:         ( IIT JEE 1995 )

  1. Are parallel to each other
  2. Are diverging
  3. Make an angle 2 [sin-1 (0.72) – 30o] with each other
  4. Make an angle 2 sin-1 (0.72) with each other

4.The focal lengths of the objective and the eyepiece of a compound microscope are 2.0 cm and 3.0 cm respectively.  The distance between the objective and the eyepiece is 15.0 cm.  The final image formed by the eyepiece is at infinity. The two lenses are thin.  The distance in cm of the object and the mage produced by the objective, measured form the objective lens, are respectively:    ( IIT JEE 1995 )

  1. 2.4 and 12.0
  2. 2.4 and 15.0
  3. 2.0 and 12.0
  4. 2.0 and 3.0

5.A diminished image of an object is to be obtained on a screen 1.0 m from it.  This can be achieved by appropriate placing: ( IIT JEE 1995 )

  1. A concave mirror of suitable focal length
  2. A convex mirror of suitable focal length
  3. A convex lens of focal length less than 0.25 m
  4. A convex lens of suitable focal length

6. Consider Fraunhofer diffraction pattern obtained with a single slit illuminated at normal incidence.  At the angular position of the first diffraction minimum the phase difference (in radian) between the wavelets from the opposite edges of the slit is – ( IIT JEE 1995 )

  1. π/4
  2. π/2
  3. π

7.An eye specialist prescribes spectacles having combination of convex lens of focal length 40 cm in contact with a concave lens of focal length 25 cm. The power of this lens combination in diopters is  – ( IIT JEE 1997 )

  1. + 1.5
  2. – 1.5
  3. + 6.67
  4. -6.67

8. A real image of a distant object is formed by a planoconvex lens on its principal axis.  Spherical aberration – ( IIT JEE 1998 )

  1. Is absent
  2. Is smaller if the curved surface of the lens faces the object
  3. Is smaller if the plane surface of the lens faces the object
  4. Is the same whichever side of the lens faces the object.

9. A parallel monochromatic beam of light is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam.  At the first minimum of the diffraction pattern, the phase difference between the rays coming from the two edges f the slit is – ( IIT JEE 1998 )

  1. Zero
  2. π/2
  3. π

10.A concave mirror is placed on a horizontal table with its axis directed vertically upwards.  Let O be the pole of the mirror and C its centre of curvature.  A point object  is placed at C.  It has a real image, also located at C.  If the mirror is now filled with water, the image will be – ( IIT JEE 1998 )

  1. Real and will remains at C
  2. Real and located at a point between C and µ
  3. Virtual and located at a point between C and O
  4. Real and located at a point between C and O

11.A spherical surface of radius of curvature R, separates air (refractive index 1.0) from glass (refractive index 1.5).  The centre of curvature is in the glass.  A point object P placed in air is found to have a real image Q in the glass.  The line PQ cuts the surface at a point O and PO = OQ.  The distance PO is equal to –  ( IIT JEE 1998 )

  1. 5 R
  2. 3 R
  3. 2 R
  4. 1.5 R

12.Yellow light is used in a single slit diffraction experiment with slit width of 0.6 mm.  If yellow light is replaced by X-rays, then the observed pattern will reveal –  ( IIT JEE 1999 )

  1. That the central maximum is narrower
  2. More number of fringes
  3. Less number of fingers
  4. No diffraction pattern

13. A thin slice is cut out of a glass cylinder along a place parallel to its axis. The slice is placed on a flat plate .  The observed interference fringes from this combination shall be  –  ( IIT JEE 1999 )

  1.  Straight
  2. Circular
  3. Equally spaced
  4. Having fringe spacing which increases as we go outwards

14. A concave lens of glass, refractive index 1.5 has both surfaces of same radius of curvature R.  On immersion in a medium of refractive index 1.75, it will behave as a –  ( IIT JEE 1999 )

  1. Convergent lens of focal length 3.5 R
  2. Convergent lens of focal length 3.0 R
  3. divergent lens of focal length 3.5 R
  4. divergent lens of focal length 3.0 R

15. In a compound microscope, the intermediate image is  –  ( IIT JEE 2000 )

  1. Virtual, erect and magnified
  2. Real, erect and magnified
  3. Real, inverted and magnified
  4. Virtual, erect and reduced

16. A hollow double concave lens is made of very thin transparent material. It can be filled with air or either of two liquids L1 and L2 having refracting indices n1 and n2 respectively (n2 > n1 > 1).  The lens will diverge a parallel beam of light if it is filled with – ( IIT JEE 2000 )

  1. Air and placed in air
  2. Air and immersed in L1
  3. L1 and immersed in L2
  4. L2 and immersed in L1

17.A diverging beam of light from a point source S having divergence angle a falls symmetrically on a glass slab as shown. The angles of incidence of the two extreme rays are equal.  If the thickness of the glass slab is t and its refractive index is n, then the divergence angle of the emergent beam is –  ( IIT JEE 2000 )

  1. Zero
  2. Sin-1 (1/n)
  3. 2sin-1 (1/n)

18. In a double slit experiment instead of taking slits of equal widths, one slit is made twice as wide as the other, then in the interference pattern -( IIT JEE 2000 )

  1. The intensities of both the maxima and the minima increases
  2. The intensity of the maxima increases and the minima has zero intensity
  3. The intensity of maxima decreases and that of minima increases
  4. The intensity of maxima decreases and the minima has zero intensity

19.A point source of light B, placed at a distance L in front of the centre of plane mirror of width d, hangs vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror at a distance 2L from it as shown. The greatest distance over which he can see the image of the light source in the mirror is –  ( IIT JEE 200 )

  1. d/2
  2. d
  3. 2d
  4. 3d

20. A rectangular glass slab ABCD of refractive index n1 is immersed in water of refractive index n2 (n1 > n2).  A ray of light is incident at the surface AB of the slab as shown.  The maximum value of the angle of incidence amax, such that the ray comes out only from the other surface CD, is given by –   ( IIT JEE 2000 )

  1. Sin-1 [n1/n2 cos (sin-1 n2/n1)]
  2. Sin-1 [n1 cos (sin-1 1/n2)]
  3. Sin-1 (n1/n2)
  4. Sin-1 (n2/n1)

21. Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen.  The phase difference between the beam is p/2 at point A and p at point B.  then the difference between resultant intensities at A and B is –   ( IIT JEE 2001 )

  1. 2I
  2. 4I
  3. 5I
  4. 7I

22. In a Young’s double slit experiment, 12 fringes are observed to be formed in a certain segment of the screen when light of wavelength 600 nm is used.  If the wavelength of light is changed to 400 nm, number of fringes observed in the same segment of the screen is given by- ( IIT JEE 2001 )

  1. 12
  2. 18
  3. 24
  4. 30

23. A ray if light passes through four transparent media with refractive indices μ1, μ2, μ3 and μ4 .  The surfaces of all media are parallel.  If the emergent ray CD is parallel to the incident ray AB, we must have –       ( IIT JEE 2001 )

  1. μ1 = μ2
  2. μ2 = μ3
  3. μ3 = μ4
  4. μ4 = μ1

24. A given ray of light suffers minimum deviation in an equilateral prism P.  Additional prism Q and R of identical shape and of the same material as P are now added as shown in the figure.  The ray will suffer-    ( IIT JEE 2001 )

  1. Greater deviation
  2. No deviation
  3. Same deviation as before
  4. Total internal reflection

25. An observer can see though a pin-hole the top end of a thin rod of height h, placed as shown in the figure.  The beaker height is 3h and its radius h.  When the beaker is filled with a liquid up to a height 2h, he can see the lower end of the rod.  Then the refractive index of the liquid is:

  1. 5/2                                                  ( IIT JEE 2002 )
  2. √5/2
  3. √3/2
  4. 3/2

26.In the ideal double-slit experiment, when a glass-plate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wavelength K), the intensity at the position where the central maximum occurred previously remains unchanged.  The minimum thickness of the glass-plate is –                    ( IIT JEE 2002 )

  1. 2λ/3
  2. λ/3
  3. λ

27. Two plane mirror A and B are aligned parallel to each other, as shown in the figure.  A light ray is incident at an angled 30o at a point just inside one end of A.  The plane of incident coincides with the plane of the figure.  The maximum number of times the ray undergoes reflections (including the first one) before it emerges out is:

  1. 28
  2. 30
  3. 32
  4. 34

28.In the adjacent diagram, CP represent a wavefront and AO and BP, the corresponding two rays.  Find the condition onq for constructive interference at P between the ray BP and reflected ray OP:

  1. Cos θ = 3λ/2d
  2. Cos θ =λ/4d
  3. Sec θ – cos θ = λ/d
  4. Sec θ – cos θ = 4λ/d

29. The size of the image of an object, which is at infinity, as formed by a convex lens of focal length 30 cm is 2 cm.  If a concave lens of focal length 20 cm is placed between the convex lens and the image at a distance of 26 cm from the convex lens, calculate the new size of the image.

  1. 1.25 cm
  2. 2.5 cm
  3. 1.05 cm
  4. 2 cm

30. A ray of light is incident at the glass-water interface at an angle I, it emerges finally parallel to the surface of water, then the value of μg would be:

  1. (4/3) sin i
  2. 1/sin i
  3. 4/3
  4. 1

31 .White light is incident on the interface of glass and air as shown in the figure.  If green light is just totally internally reflected then the emerging ray in air contains.

  1. Yellow, orange, red
  2. Violet, indigo, blue
  3. All colours
  4. All colours except green

32. A ray of light is incident on an equilateral glass prism placed on a horizontal table.  For minimum deviation which of the following is true?

  1. PQ is horizontal
  2. QR is horizontal
  3. RS is horizontal
  4. Either PQ or RS is horizontal

33. A point object is placed at the centre of a glass sphere of radius 6 cm and refractive index 1.5.  The distance of the virtual image from the surface of the sphere is –  ( IIT JEE 2004 )

  1. 2 cm
  2. 4 cm
  3. 6 cm
  4. 12 cm

34. In a YDSE bi-chromatic light of wavelengths 400 nm and 560 nm are used.  The distance between the sits is 0.1 mm and the distance between the plane of the slits and the screen is 1 m.  The minimum distance between two successive regions of complete darkness is – ( IIT JEE 2004 )

  1. 4 mm
  2. 5.6 mm
  3. 14 mm
  4. 28 mm

35. In Young’s double slit experiment intensity at a point is (1/4) of the maximum intensity.  Angular position of this points is ( IIT JEE 2004 )

  1. Sin-1 (λ/d)
  2. Sin-1 (λ/2d)
  3. Sin-1 (xλ/3d)
  4. Sin-1 (λ/4d)

36. A container is filled with water (m = 1.33) upto 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 – ( IIT JEE 2004 )

  1. 10 cm
  2. 15 cm
  3. 20 cm
  4. 25 cm

37. A convex lens is in contact with concave lens.  The magnitude of the ratio of their focal length is 2/3.  Their equivalent focal length is 30 cm.  What are their individual focal lengths?   ( IIT JEE 2005 )

  1. -75, 50
  2. -10, 15
  3. 75, 50
  4. -15, 10

38. A point object is placed at distance of 20 cm from a thin planoconvex lens of focal length 15 cm.  The place surface of the lens is now silvered.  The image created by the system is at –  ( IIT JEE 2006 )

  1. 60 cm to the left of the system
  2. 60 cm to the right of the system
  3. 12 cm to the left of the system
  4. 12 cm to the right of the system

39. The graph between object distance u and image distance v for a lens is given below.  The focal length of the lens is – ( IIT JEE 2006 )

  1. 5 ± 0.1
  2. 5 ± 0.05
  3. 0.5 ± 0.1
  4. 0.5 ± 0.05

40. A biconvex lens of focal length f forms a circular image of radius r of sun in focal plane.  Then which option is correct? ( IIT JEE 2006 )

  1. πr2 f
  2. πr2f 2
  3. If lower half part is converted by black sheet, then area pf the image is equal to πr2/ 2
  4. If f  is doubled, intensity will increase

41. A ray of light traveling in water is incident on its surface open to air.  The angle of incidence is θ, which is less than the critical angle.  Then there will be – ( IIT JEE 2007 )

  1. Only a reflected ray and no refracted ray
  2. Only a refracted ray and no reflected ray
  3. A reflected ray and a refracted ray and the angle between them would be less than 180o – 2θ
  4. A reflected ray and a refracted ray and the angle between them would be greater than 180o – 2θ

42. In an experiment to determine the focal length ( f ) of a concave mirror by the u-v method, a student places the object pin A on the principal axis at a distance x from the pole P.  The student looks at the pin and its inverted image from a distance keeping his /her eye in line with PA.  When the student shifts his/her eye towards left, the image appears to the right of the object pin.  Then – ( IIT JEE 2007 )

  1. X < f
  2. f < x < 2f
  3. X = 2 f
  4. X > 2 f

OBJECTIVE QUESTIONS – ( More than one options are correct ) –

1. In an interference arrangement similar to Young’s double-slit experiment, the slits S1 and S2 are illuminated with coherent microwave sources, each of frequency106 Hz.  The sources are synchronized to have zero phase difference.  The slits are separated by a distance d = 150.0m.  The intensity I (θ) is measured as a function of θ, where θ is defined as shown.  If I0 is the maximum intensity, then I (θ) for 0 ≤θ ≤ 90o is given by – ( IIT JEE 1995 )

  1. I(θ) = I0/ 2 for θ = 30o
  2. I(θ) = I0/ 4 for θ = 90o
  3. I(θ) = I0 for θ = 0o
  4. I(θ) is constant for all values of θ

2.Which of the following form(s) a virtual and erect image for all positions of the object? ( IIT JEE 1996 )

  1. Convex lens
  2. Concave lens
  3. Convex mirror
  4. Concave mirror

3. A ray of light traveling in a transparent medium falls on a surface separating the medium from air at an angle of incident 45°.  the ray undergoes total internal reflection.  If n is the refractive indeed of the medium with respect to air, select the possible value(s) of n from the following – ( IIT JEE 1998 )

  1. 1.3
  2. 1.4
  3. 1.5
  4. 1.6

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