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Ray Optics and Optical Instruments Quiz-16

Dear Readers,


As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background. Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. .

Q1.  The unit of focal power of a lens is
  •   Watt
  •   Horse power
  •   Dioptre
  •   Lux
Solution
Dioptre

Q2. Two lenses of power – 15D and +5 D are in contact with each other. The focal length of the combination is
  •   -20 cm
  •   -10 cm
  •   +20 cm
  •   +10 cm
Solution
Power of lens is reciprocal of its focal length. Power of combined lens is ๐‘ƒ = ๐‘ƒ₁ + ๐‘ƒ₂ = −15 + 5 = −10 ๐ท ∴ ๐‘“ = 1/๐‘ƒ = 100/−10 ๐‘๐‘š ๐‘“ = −10 ๐‘๐‘š

Q3.   A telescope of diameter 2๐‘š uses light of wavelength 5000 โ„ซ for viewing stars. The minimum angular separation between two stars whose image is just resolved by this telescope is
  •   4 × 10-4๐‘Ÿ๐‘Ž๐‘‘
  •   0.25 × 10-6 ๐‘Ÿ๐‘Ž๐‘‘
  •   0.31 × 10-6 ๐‘Ÿ๐‘Ž๐‘‘
  •   5.0 × 10-3๐‘Ÿ๐‘Ž๐‘‘
Solution
Minimum angular separation ∆๐œƒ = 1/๐‘….๐‘ƒ. = 1.22 ๐œ†/๐‘‘ = (1.22 × 5000 × 10−10)/ 2 = 0.3 × 10-6๐‘Ÿ๐‘Ž๐‘‘

Q4. Let the ๐‘ฅ − ๐‘ง plane be the boundary between two transparent media. Medium 1 in ๐‘ง ≥ 0 has a refractive index of √2 and medium 2 with ๐‘ง < 0 has a refractive index of√3. A ray of light in medium 1 given by the vector๐€ = 6√3รฎ + 8√3ฤต – 10 ๐’Œ ̂ is incident on the plane of separation. The angle of refraction in medium 2 is
  •   45°
  •   60°
  •   75°
  •   30°
Solution
As refractive index for ๐‘ง > 0 and ๐‘ง ≤ 0 is different ๐‘‹ − ๐‘Œ plane should be boundary between two media Angle of incidence,
 cos๐‘– = | | ๐ด๐‘ง/ √๐ดx2 + ๐ดy2 + ๐ดz2 | | = 1/2 ∴ ๐‘– = 60° 
From Snell’s law 
sin๐‘–/sin๐‘Ÿ = √3/2 ⇒ ๐‘Ÿ = 45°

Q5. A light beam is travelling from Region I to Region IV (refer figure). The refractive index in Region I, II, III and IV are ๐‘›o, ๐‘›o/2 , ๐‘›o/6 and ๐‘›o/8 , respectively. The angle of incidence ฮธ for which the beam just misses entering Region IV is

 
  •   sin−1 (3/4 )
  •   sin−1 (1/8 )
  •   sin−1 (1/4 )
  •   sin−1 (1/3 )
Solution
  Critical angle from region III to region IV sinฮธc= (๐‘›o/8)/ (๐‘›o/6) = 3/4 Now applying Snell’s law in region I and region III ๐‘›o sinฮธ = ๐‘›o/6 sinฮธc Or sinฮธ = 1/6 sinฮธc = 1/6 (3/4 ) = 1/8 ∴ ฮธ = sin-1( 1/8 ) .

Q6.  A rectangular glass slab ๐ด๐ต๐ถ๐ท, of refractive index ๐‘›1, is immersed in water of refractive index ๐‘›2(๐‘›1 > ๐‘›2). A ray of light in incident at the surface ๐ด๐ต of the slab as shown. The maximum value of the angle of incidence ๐›ผ๐‘š๐‘Ž๐‘ฅ such that the ray comes out only from the other surface ๐ถ๐ท is given by 

 
  •   sin-1 [ (๐‘›1/๐‘›2) cos(sin-1 ๐‘›2/๐‘›1 )]
  •   sin-1 [๐‘›1 cos(sin-1 1/๐‘›2 )]
  • sin-1 (๐‘›1/๐‘›2)
  •   sin-1 ( ๐‘›2/๐‘›1 )
Solution
Ray comes out from ๐ถ๐ท, means rays after refraction from ๐ด๐ต get, total intensity reflected at ๐ด๐ท 
                                                      


 – ๐‘›1/๐‘›2 = sin๐›ผmax/sin๐‘Ÿ1 ⇒ ๐›ผmax = sin-1[ ๐‘›1/๐‘›2 sin๐‘Ÿ1] …(i) 
 Also ๐‘Ÿ1 + ๐‘Ÿ2 = 90° 
⇒ ๐‘Ÿ1 = 90 − ๐‘Ÿ2 = 90 − ๐ถ ⇒ ๐‘Ÿ1 = 90 − sin-1 ( 1/2๐œ‡1) ⇒ ๐‘Ÿ1 
= 90 − sin-1 (๐‘›2/๐‘›1) …(ii) 
Hence from equation (i) and (ii)
 ๐›ผmax= sin-1 [ ๐‘›1/๐‘›2 sin(90 − sin-1 ๐‘›2/๐‘›1 )] 
= sin-1[(๐‘›1/๐‘›2)cos(sin-1 (๐‘›2/๐‘›1)]


Q7.A vessel of depth 2d ๐‘๐‘š is half filled with a liquid of refractive index ๐œ‡1 and the upper half with a liquid of refractive index ๐œ‡2. The apparent depth of the vessel seen perpendicularly is
  •   ๐‘‘ ( (๐œ‡1๐œ‡2)/(๐œ‡1+ ๐œ‡2) )
  •   ๐‘‘ ( 1/๐œ‡1 + 1/๐œ‡2 )
  •   2๐‘‘ ( 1/๐œ‡1 + 1/๐œ‡2 )
  •   2๐‘‘ ( 1/๐œ‡1๐œ‡2 )
Solution

โ„Ž′ = ๐‘‘1/๐œ‡1 + ๐‘‘2/๐œ‡2 = ๐‘‘ ( 1/๐œ‡1 + 1/๐œ‡2 )


Q8. 1% of light of a source with luminous intensity 50 ๐‘๐‘Ž๐‘›๐‘‘๐‘’๐‘™๐‘Ž is incident on a circular surface of radius 10 ๐‘๐‘š. The average illuminance of surface is
  •   100 ๐‘™๐‘ข๐‘ฅ
  •   200 ๐‘™๐‘ข๐‘ฅ
  •   300 ๐‘™๐‘ข๐‘ฅ
  •   400 ๐‘™๐‘ข๐‘ฅ
Solution
๐œ™= 4๐œ‹๐ฟ = 200 ๐œ‹ ๐‘™๐‘ข๐‘š๐‘’๐‘› So ๐ผ = ๐œ™/100 ๐ด = 200 ๐œ‹/100×๐œ‹๐‘Ÿ2 = 2/(0.1)2 = 200 ๐‘™๐‘ข๐‘ฅ


Q9. If both the object and image are at infinite distance from a refracting telescope its magnifying power will be equal to

  •   The sum of the focal lengths of the objective and the eyepiece
  •   The different of the focal lengths of the two lenses
  •   The ratio of the focal length of the objective and eyepiece
  •   The ratio of the focal length of the eyepiece and objective
Solution
In case of a telescope if object and final image are at infinity then ๐‘š = ๐‘“๐‘œ/๐‘“๐‘’

Q10.  One side of a glass slab is silvered as shown. A ray of light is incident on the other side at angle of incidence ๐‘– = 45°. Refractive index of glass is given as 1.5. the deviation of the ray of light from its initial path when it comes out of the slab is


  •   90°
  •   180°
  •   120°
  •   45°
Solution
From the figure it is clear that the angle between incident ray and emergent ray is 90° 



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RAY OPTICS AND OPTICAL INSTRUMENTS QUIZ 16
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