DIRECTIONS (Q. No. 14) : Each question contains statements given in two columns, which have to be matched. The statements in ColumnI are labelled A, B, C and D, while the statements in ColumnII are labelled p, q, r and s. Any given statement in ColumnI can have correct matching with ONE OR MORE statement(s) in ColumnII. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example :
If the correct matches are Ap, s and t; Bq and r; Cp and q; and Ds then the correct darkening of bubbles will look like the given.
Q.1. A simple telescope used to view distant objects has eyepiece and objective lens of focal lengths fe and fo, respectively. Then (2006  6M)
Column I  Column II 
(A) Intensity of light received by lens  (p) Radius of aperture 
(B) Angular magnification  (q) Dispersion of lens 
(C) Length of telescope  (r) Focal length of objective lens and eyepiece lens 
(D) Sharpness of image  (s) Spherical aberration 
Ans. Ap; Br; Cr; Dp, q, s
Solution. (A) → (p).
More the radius of aperture more is the amount of light entering the telescope.
(B) → (q).
Depends on dispersion of lens, spherical aberration and radius of aperture.
Q.2. An optical component and an object S placed along its optic axis are given in Column I. The distance between the object and the component can be varied. The properties of images are given in Column II. Match all the properties of images from Column II with the appropriate components given in Column I. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS.
Column I  Column II 
(p) real image  
(q) virtual image  
(r) magnified image  
(s) image at infinity 
Ans. Ap, q, r, s; Bq; Cp, q, r, s; Dp, q, r, s
Solution. Ap, q, r, s l
Q.3. ColumnI shows four situations of standard Young’s double slit arrangement with the screen placed far away from the slits S_{1} and S_{2}. In each of these cases S_{1}P_{0} = S_{2}P_{0}, S_{1}P_{1}S_{2}P_{1 }= λ/4 and S_{1}P_{2}– S_{2}P_{2 }= λ /3, where λ is the wavelength of the light used. In the cases B, C and D, a transparent sheet of refractive index μ and thickness t is pasted on slit S_{2}. The thicknesses of the sheets are different in different cases. The phase difference between the light waves reaching a point P on the screen from the two slits is denoted by d (P) and the intensity by δ(P). Match each situation given in ColumnI with the statetment(s) in ColumnII valid for that situation.
ColumnI  ColumnII 
(p) δ(P_{0}) = 0  
(q) δ (P_{1}) = 0  
(r) I (P_{1}) = 0  
(s) I(P_{0}) > I (P_{1})  
(t) I(P_{2}) > I (P_{1}) 
Ans. Ap, s; Bq; Ct; Dr, s, t
Solution.
For path difference λ /4 , phase difference is π /2 .
For path difference λ/3 , phase difference is 2π /3 .
Here, S_{1}P_{0}  S_{2}P_{0 }= 0
∴ δ (P_{0}) = 0
Therefore, (p) matches with (A).
The path difference for P_{1} and P_{2} will not be zero. The intensities at P_{0} is maximum.
Therefore, (s) matches with (A).
(B)
Therefore, q match with (B)
(C)
∴ I (P_{2}) > I (P_{1})
(t) matches (C).
(D)
(r), (s), (t) matches (D).
Q.4. Two transparent media of refractive indices μ_{1} and μ_{3} have a solid lens shaped transparent material of refractive index μ_{2} between them as shown in figures in Column II. A ray traversing these media is also shown in the figures. In Column I different relationships between μ_{1}, μ_{2}, and μ_{3} are given. Match them to the ray diagrams shown in Column II. (2010)
Column I  Column II 
(A) μ_{1} < μ_{2}  
(B) μ_{1} > μ_{2}  
(C) μ_{2} = μ_{3}  
(D) μ_{2} > μ_{3}  
Ans. Ap, r; Bq, s, t; Cp, r, t; Dq, s
Solution. (a) When μ_{1 }< μ_{2}, the ray of light while entering the lens will bend towards the normal. Therefore p, r are the correct options
(B) When μ_{1}> μ_{2}, the ray of light while entering the lens will bend away from the normal. Therefore q,s,t are the correct options.
(C) When μ_{2} = μ_{3}, the ray of light while coming out from the lens does not deviate from its path. Therefore p,r,t are the correct option.
(D) μ_{2}> μ_{3}, the ray of light coming out of the lens deviates away from the normal. Therefore q,s are the correct options.
DIRECTION (Q. No. 5 & 6) Following question has matching lists. The codes for the lists have choices (a), (b), (c) and (d) out of which ONLY ONE is correct.
Q.5. A right angled prism of refractive index μ_{1} is placed in a rectangular block of refractive index μ_{2}, which is surrounded by a medium of refractive index μ_{3}, as shown in the figure. A ray of light 'e' enters the rectangular block at normal incidence. Depending upon the relationships between μ_{1}, μ_{2} and μ_{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 Adv. 2013)
List I  List II 
P. e → f  1. μ_{1} > 2μ_{2} 
Q. e → g  2. μ_{2} > μ_{1} and μ_{2} > μ_{3} 
R. e → h  3. μ_{1} = μ_{2} 
S. e → i 
Ans. (d)
Solution. e → f. For the ray to bend towards the normal at the prism surface μ_{2} > μ_{1}. The ray then moves away from the normal when it emerges out of the rectangular block. Therefore μ_{2} > μ_{3}.
e → g. As there is no deviation of the ray as it emerges out of the prism, μ_{2} = μ_{1}.
e → h. As the ray emerges out of prism, it moves away from the normal. Therefore μ_{2} < μ_{1}. As the ray moves away from the normal as it emerges out of the rectangular block, therefore μ_{2} > μ_{3}.
e → i. At the prism surface, total internal reflection has taken place. For this sin
∴ m1 > √2 μ_{2}. (d) is the correct option.
Q.6. Four combinations of two thin lenses are given in ListI. The radius of curvature of all curved surfaces is r and the refractive index of all the lenses is 1.5. Match lens combinations in ListI with their focal length in ListII and select the correct answer using the code given below the lists. (JEE Adv. 2014)
List  I  List  II 
1. 2r  
2. r/2  
3. – r3. – r  
4. r 
Codes:
(a) P1, Q2, R3, S4
(b) P2, Q4, R3, S1
(c) P4, Q1, R2, S3
(d) P2, Q1, R3, S4
Ans. (b)
Solution.
For the combination
∴ f = 2r
For the combination
∴ f = r
Similarly, we can either find or do not find the remaining options (b) is the correct option.
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