IIT JEE Physics : Optics previous 2011 – 2014 questions & answers.

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

 

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1 . A transparent thin film of uniform thickness and refractive index {n}_{1}=1.4 is coated on the convex spherical surface of radius R at one end of a long solid glass cylinder of refractive index {n}_{2}=1.5 as shown in the below figure. The rays of light parallel to the axis of the cylinder traversing through the film from air to glass get focused at distance {f}_{1} from the film, while rays of light traversing from glass to air get focused at distance {f}_{2} from the film. Then   (JEE Advanced 2014)

(a) \mid{{f}_{{1}}}\mid=3R
(b) \mid{{f}_{{1}}}\mid=2.8 R
(c) \mid{{f}_{{2}}}\mid=2R
(d) \mid{{f}_{{2}}}\mid=1.4 R

Solution

(a, (c) As film has {R}_{1}={R}_{2}=R (say), its focal length is
\frac{1}{f}=\left({{\mu }_{{1}}-1}\right)\left({\frac {1}{R}-\frac {1}{R}}\right)

 

If \frac{1}{f}=0 or f=\infty it will not cause refraction. The refraction takes place at air and glass cylinder. It is given that u=\infty
\frac {-{\mu }_{{1}}}{u}+\frac {{\mu }_{{2}}}{\nu }=\frac {{\mu }_{{2}}-{\mu }_{{1}}}{R}
-\frac {{\mu }_{{1}}}{\infty }+\frac {1\cdot 5}{{f}_{{1}}}=\frac{1.5-1}{R} or {f}_{1}=3R
For glass-air refraction, we have
-\frac {{\mu }_{{2}}}{u}+\frac {{\mu }_{{1}}}{v}=\frac {{\mu }_{{1}}-{\mu }_{{2}}}{R}
-\frac {{\mu }_{{2}}}{\infty }+\frac {1}{{b}_{{2}}}=\frac{1-1.5}{R}
of {f}_{2}=-2R or \mid{{b}_{{2}}}\mid=2R.

 

2 . A point source S is placed at the bottom of a transparent block of height 10 mm and refractive index 2.72. It is immersed in a lower refractive index liquid emerging from the block to the liquid forms a circular bright spot of diameter 11.54 mm on the top of the block. The refractive index of the liquid is       (JEE Advanced 2014)

(a) 1.21
(b) 1.30
(c) 1.36
(d) 1.42

Solution

(c)

In \Delta{SAB}
{\tan{\theta }}=\frac{AB}{AS}=\frac {\left({\frac {11\cdot 54}{2}}\right)}{10}
\Rightarrow{\tan{\theta }}=0.577
\Rightarrow{\theta}_{c}={30}^{\circ}
Using \frac {\sin{{\theta }_{{c}}}}{\sin{90}}=\frac {{\mu }_{{l}}}{{\mu }_{{b}}}, we get
{\sin{{\theta }_{{c}}}}=\frac {{\mu }_{{l}}}{{\mu }_{{b}}} or {\mu}_{b}\times{\sin{30}}=2.72\times\frac{1}{2}=1.36

 

3 . Four combination of two thin lenses are given in List I. The radius of curvature of all curved surfaces is r and the refractive index of all curved surfaces is r and the refractive index of all the lenses is 1.5. Match lens combination in List I with their focal length in List II and select the correct answer using the code given below the lists.

List I            List II
P.                1. 2r

Q.                2. \frac{r}{2}

R.                 3. -r

S.                 4. r

Codes:
(a) P-1, Q-2, R-3, S-4
(b) P-2, Q-4, R-3, S-1
(C) P-4, Q-1, R-2, S-3
(D) P-2, Q-1, R-3, S-4

Solution

(b) Applying lens maker’s formula,
\frac{1}{f}=(\mu-1)(\frac {1}{{R}_{{1}}}-\frac {1}{{R}_{{2}}})
\frac{1}{{f}_{eff}}=\frac{1}{{f}_{1}}+\frac{1}{{f}_{2}}

Case P: Here ,
\frac{1}{f}=0.5(\frac {1}{r}-\frac {1}{-r})
and r=f. Therefore
\frac{1}{{f}_{eff}}=\frac{1}{r}+\frac{1}{r}\Rightarrow{f}_{eff}=\frac{r}{2}
Therefore, (P)\rightarrow(2)
Case Q: We have
\frac{1}{f}=0.5(\frac {1}{\infty}-\frac {1}{-r})=\frac{1}{2r}
\frac{1}{{f}_{eff}}=\frac{1}{2r}+\frac{1}{2r}\Rightarrow{f}_{eff}={r}
Therefore, (Q)\rightarrow(4)
Case R: We have
\frac{1}{f}=0.5(\frac {1}{\infty}-\frac {1}{r})=\frac{-1}{2r}
\frac{1}{{f}_{eff}}=-\frac{1}{2r}-\frac{1}{2r}\Rightarrow{f}_{eff}={-r}
Therefore, (R)\rightarrow(3)
\frac{1}{{f}_{eff}}=\frac{1}{r}-\frac{1}{2r}=\frac{1}{2r}\Rightarrow{f}_{eff}={2r}
Therefore, (S)\rightarrow(1).

4 . The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real and one-third the size of the object. The wave length of light inside the lens is 2/3 times the wavelength in free space. The radius of the curved surface of the lens is       (JEE Advanced 2013)
(a) 1 m
(b) 2 m
(c) 3 m
(d) 6 m

Solution
(c) The refractive index is
\mu=\frac{c}{v}=\frac {f{\lambda }_{{air}}}{f{\lambda }_{{med}}}=\frac {{\lambda }_{{air}}}{{\lambda }_{{med}}}=\frac{3}{2}
Now, v=+8m. As image is real,
m=\frac{-1}{3}\Rightarrow{u}=\frac {8}{-\frac {1}{3}}=-24 m
Using the lens formula,we get
-\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\Rightarrow\frac{1}{f}=\frac{1}{v}-\frac{1}{u}
We have, u=-24 m and v=+8 m. Hence,
\frac{1}{f}=\frac{1}{8}+\frac{1}{24}=\frac{4}{24}
\Rightarrow{f}=6 m
Using lensmaker’s formula,
\frac{1}{f}=(\mu-1)[\frac{1}{R}-\frac{1}{\infty}]
f=\frac{R}{(mu-1)}=\frac{R}{1.5-1}
6=\frac{R}{0.5}
\Rightarrow{R}=3 m

 

5  . A right angled prism of refractive index {\mu}_{1} is placed in a rectangular block of refractive index {\mu}_{2}, which is surrounded by a medium of refractive index {\mu}_{3}, as shown in the figure. A ray of light e enters the rectangular block at normal incidence. Depending upon the relashionships between {\mu}_{1},{\mu}_{2}, and {\mu}_{3} it takes one of the four possible paths ‘ef’,’eg’,’eh’ or ‘ei’.

 

Match the paths in List I with conditions of refractive indices in List II and select the correct answer using the codes given below the lists: (JEE Advanced 2013)
LIst I                                               List II
(P) {e}\rightarrow{f}            (1) {\mu}_{1}>\sqrt{{2}}{\mu}_{2}
(Q) e\rightarrow{g}              (2) {\mu}_{2}>{\mu}_{1} and {\mu}_{2}>{\mu}_{3}
(R) e\rightarrow{h}               (3) {\mu}_{1}={\mu}_{2}
(Q) e\rightarrow{i}                (4) {\mu}_{2}<{\mu}_{1}<\sqrt{{2}}{\mu }_{{2}} and {\mu}_{2}>{\mu}_{3}

Codes:

(P)    (Q)    (R)    (S)
(a)   2      3      1      4
(b)   1      2      4      3
(c)   4      1      2      3
(d)   2      3      4      1

Solution

(d)
For path {e}\rightarrow{f};{\mu}_{2}>{\mu}_{1} and {\mu}_{3}<{\mu}_{2} Thus, the correct mapping is (P)\rightarrow(2)
For path e\rightarrow{g}:{\mu}_{1}={\mu}_{2} (no bending). Hence , the correct mapping is (Q)\rightarrow(3)
For e\rightarrow{h}:{\mu}_{2}<{\mu}_{1} and {\mu}_{3}<{\mu}_{2} Also, {\mu}_{1}<\sqrt{{2}}{\mu }_{{2}} (no total internal reflection). Hence , the correct mapping is (R)\rightarrow(4)
For e\rightarrow{i} the total internal reflection occurs, that is 45^{\circ}> \frac {{\mu }_{{2}}}{{\mu }_{{1}}}\Rightarrow {\mu }_{{1}}> \sqrt{{2}}{\mu }_{{2}}. Therefore, the correct mapping is (S)\rightarrow(1)

6 . A ray of light travelling in the direction \frac{1}{2}\left({\hat{i}+3\hat{J}}\right) is incident on a plane mirror. After reflection, it travels along the direction \frac{1}{2}\left({\hat{i}-3\hat{J}}\right). The angle of incidence is         (JEE Advanced 2013)
(a) {30}^{\circ}
(b) {45}^{\circ}
(c) {60}^{\circ}
(d) {75}^{\circ}

(a)

Vectors \vec{A} and \vec{B} are orginating from O have angle \left({180^{\circ}-2\theta }\right) between them. Using \vec{A}\cdot \vec{B}=AB{\cos{\theta }} we get
{\cos{\left({180^{\circ}-2\theta }\right)}}=\frac{\left({\frac {\hat{i}}{2}+\frac {\sqrt{{3}}}{2}\hat{j}}\right)\cdot\left({\frac {\hat{i}}{2}-\frac {\sqrt{{3}}}{2}\hat{j}}\right)}{\mid{\frac {\hat{i}}{2}+\frac {\sqrt{{3}}}{2}\hat{j}}\mid\mid{\frac {\hat{i}}{2}-\frac {\sqrt{{3}}}{2}\hat{j}}\mid}‘
=\frac{\left({\frac {1}{4}-\frac {3}{4}}\right)}{\left({\frac {1}{4}+\frac {3}{4}}\right)\left({\frac {1}{4}+\frac {3}{4}}\right)}=-\frac{1}{2}
{\cos{\left({180^{\circ}-2\theta }\right)}}={120}^{\circ}
\theta={30}^{\circ}

7 . A bi-convex lens is formed with two plano-convex lenses as shown in the figure. Refractive index n of the first lens is 1.5 and that of the second lens is 1.2. Both curved surface are of the same radius of curvature R = 14 cm. For this bi-convex lens, for an object distance of 40cm, the image distance will be    (JEE Advanced 2012)

(a) -280.0cm
(b) 40.0 cm
(c) 21.5 cm
(d) 13.3 cm

Solution

(b) Considering the biconvex lens, for an object distance of 40 cm, the image distance is found out as follows:
{P}_{T}=(1.5-1)(\frac{1}{14}-0)+(1.2-1)(0-\frac{1}{-14})
=\frac{0.5}{14}+\frac{0.2}{14}=\frac{1}{20}
That is, f=+20 cm. Thereforee.
\frac{1}{v}-\frac{1}{-14}=\frac{1}{20}
\Rightarrow\frac{1}{v}=\frac{1}{20}-\frac{1}{40}=\frac{1}{40}

Therefore, v = 40 cm.

Paragraph for questions 8  and  9 : 

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, \frac {\sin{{\theta }_{{1}}}}{\sin{{\theta }_{{2}}}}=\frac {{n}_{{2}}}{{n}_{{1}}}, 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=(\frac{c}{v})=\pm\sqrt{{{\epsilon }_{{r}}{\mu }_{{r}}}} where c is the speed of electromagnetic waves in vaccum, v its speed in the medium {\epsilon}_{r} and {\mu}_{r} are the relative permittivity and permeability of the medium, respectively.   (JEE Advanced 2012 )

In normal materials, both {\epsilon}_{r} and {\mu}_{r} are positive, implying positive n for the medium. When both {\epsilon}_{r} and {\mu}_{r} 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 and 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.

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

Solution

(c) Because n is negative for meta-materials .

9 . Choose the correct statement.
(a) The speed of light in the meta-material is v=c|n|
(b) The speed of light in the meta-material is v=\frac{c}{|n|}
(c) The speed of light in the meta-material is v =c.
(d) The wavelength of the light in the meta-material ({\lambda}_{m}) is given by {\lambd}_{m}={\lambda}_{air} where {\lambda}_{air} is wavelength of the light in air.

Solution
(b) Speed of light in meta-material v=\frac{c}{|n|}.

 

10 . A light ray travelling in glass medium is incident on glass-air interface at an angle of incidence \theta. The reflected (R) and transmitted (T) intensities, both as function of \theta are plotted. The correct sketch is   ( JEE Advanced 2011)

Solution
(c) As there is reflection of incident rays even if incident angle is zero. the percentage of transmission is less than 100 and high at zero angle of incidence. Due to the total internal reflection transmitted, the intensity becomes zero after the critical angle.

 

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