## RAY OPTICS AND OPTICAL INSTRUMENTS-12 Quiz

JEE Advanced Physics Syllabus can be referred by the IIT aspirants to get a detailed list of all topics that are important in cracking the entrance examination. JEE Advanced syllabus for Physics has been designed in such a way that it offers very practical and application-based learning to further make it easier for students to understand every concept or topic by correlating it with day-to-day experiences. In comparison to the other two subjects, the syllabus of JEE Advanced for physics is developed in such a way so as to test the deep understanding and application of concepts..

Q1. The table below shows object and image distances for four objects placed in front of mirrors. For which one is the image formed by a convex spherical mirror. [Positive and negative signs are used in accordance with standard sign convention
•  -7.10 cm -18.0 cm
•  -25.0 cm -16.7 cm
•  -5.0 cm +1.0 cm
•  -20.0 cm +5.71 cm
Solution

For (a) and (b), object and image are on the same side, so image is real as the object is real For (c), image is virtual but |m|>1. So concave mirror For (d), |m|less than 1 and image is also virtual, so in (d) image is formed in a convex mirror
Q2.The refractive index of a prism is 2. The prism can have a maximum refracting angle of
•  90°
•  60°
•  45°
•  90°
Solution

Critical angle Î¸_c=sin^(-1)⁡(1/Î¼) =sin^(-1)⁡(1/2)=30° If A>2Î¸_c, the ray does not emerge from the prism. So, maximum angle can be 60°

Q3.  A point source of light S is placed in front of a perfectly reflecting mirror as shown in the figure. ∑is a screen. The intensity at the centre of screen is found to be I If the mirror is removed, then the intensity at the center of screen would be  is placed in front of a perfectly reflecting mirror as shown in the figure. . The intensity at the centre of screen is found to be

•   I
•  10I/9
•  9I/10
•  2I
Solution

Let power of light source be P, then intensity at any point on the screen is due to light rays directly received from source and that due to light rays after reflection from the mirror I=P/(4Ï€a^2 )+P/(4Ï€×(3a)^2 )
When mirror is taken away,
I_1=P/(4Ï€a^2 )=9I/10

Q4.
The refracting angle of a prism is  and refractive index of the material of prism is . The angle of minimum deviation will be
•  180°-3A
•  180°+3A
•  90°-3A
•  180°-2A
Solution

From sin⁡((A+Î´_m)/2)=Î¼ sin⁡A/2 sin⁡((A+Î´_m)/2)=cos⁡A/2/sin⁡A/2 ×sin⁡A/2
(A+Î´_m)/2=Ï€/2-A/2
Î´_m=Ï€-2A=180°-2A
Q5.A point object is kept in front of a plane mirror. The plane mirror is doing SHM of amplitude 2 cm. The plane mirror moves along the -axis which is normal to the mirror. The amplitude of the mirror is such that the object is always in front of the mirror. The amplitude of SHM of the image is
•  0 cm
•  2 cm
•  4 cm
•  1 cm
Solution

Q6. An eye specialist prescribes spectacles having a combination of convex lens of focal length  in contact with a concave lens of focal length . The power of this lens combination in dioptres is
•  +1.5
•  -1.5
•  +6.67
•  -6.67
Solution

Power of convex lens P_1=100/40=2.5 D
Power of concave lens P_2=-100/25=-4 D
Now P=P_1+P_2=2.5 D-4 D=-1.5 D
Q7. The lateral magnification of the lens with an object located at two different positions  and  are  and  respectively. Then the focal length of the lens is
•  f=√(m_1 m_2 )(u_2-u_1)
•  f=√(m_1 m_2 )(u_2-u_1)
•  ((u_2-u_1))/√(m_2 m_1 )
•  ((u_2-u_1))/((m_2 )^(-1)-(m_1 )^(-1) )
Solution

For lens 1/f=1/v_1 -1/u_1 u_1/f=u_1/v_1 -1 ⇒ m_1=v_1/u_1 =f/(u_1+f)
And m_2=f/(u_2+f)
1/m_2 -1/m_1 =((u_2-u_1))/f ⇒ f=((u_2-u_1))/((m_2 )^(-1)-(m_1 )^(-1) )
Q8. In a thick glass slab of thickness , and refractive index , a cuboidal cavity of thickness ‘ ’ is carved as shown in the fig and is filled with a liquid of R.I. . The ratio , so that shift produced by this slab is zero when an observer  observes an object  with paraxial rays is
•  (n_1-n_2)/(n_2-1)
•  T(n_1-n_2)/(n_2 (n_1-1))
•  (n_1-n_2)/(n_1-1)
•  (n_1-n_2)/(n_1 (n_2-1))
Solution

Shift =(l-m)(1-1/n_1 )+m(1-1/n_2 )=0
Q9.A ray of light enters a rectangular glass slab of refractive index  at an angle of incidence . It travels a distance of 5 cm inside the slab and emerges out of the slab. The perpendicular distance between the incident and the emergent rays is
•  5√3 cm
•  5/2 cm
•  5 √(3/2) cm
•  5 cm
Solution
Q10.
 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?
•  -75, 50
•  -10, 15
•  75, 50
• -15, 10
Solution

Let focal length of convex lens is +f then focal length of concave lens would be -3/2 f. From the given condition,
1/30=1/f-2/3f=1/3f
∴ f=10 cm
Therefore, focal length of convex lens = + 10 cm and that of concave lens = -15 cm.

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