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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 𝑃 = 𝑃₁ + 𝑃₂ = −15 + 5 = −10 𝐷 ∴ 𝑓 = 1/𝑃 = 100/−10 𝑐𝑚 𝑓 = −10 𝑐𝑚

Q3.   A telescope of diameter 2𝑚 uses light of wavelength 5000 Å for viewing stars. The minimum angular separation between two stars whose image is just resolved by this telescope is
•   4 × 10-4𝑟𝑎𝑑
•   0.25 × 10-6 𝑟𝑎𝑑
•   0.31 × 10-6 𝑟𝑎𝑑
•   5.0 × 10-3𝑟𝑎𝑑
Solution
Minimum angular separation ∆𝜃 = 1/𝑅.𝑃. = 1.22 𝜆/𝑑 = (1.22 × 5000 × 10−10)/ 2 = 0.3 × 10-6𝑟𝑎𝑑

Q4. Let the 𝑥 − 𝑧 plane be the boundary between two transparent media. Medium 1 in 𝑧 ≥ 0 has a refractive index of √2 and medium 2 with 𝑧 < 0 has a refractive index of√3. A ray of light in medium 1 given by the vector𝐀 = 6√3î + 8√3ĵ – 10 𝒌 ̂ is incident on the plane of separation. The angle of refraction in medium 2 is
•   45°
•   60°
•   75°
•   30°
Solution
As refractive index for 𝑧 > 0 and 𝑧 ≤ 0 is different 𝑋 − 𝑌 plane should be boundary between two media Angle of incidence,
cos𝑖 = | | 𝐴𝑧/ √𝐴x2 + 𝐴y2 + 𝐴z2 | | = 1/2 ∴ 𝑖 = 60°
From Snell’s law
sin𝑖/sin𝑟 = √3/2 ⇒ 𝑟 = 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 𝑛o, 𝑛o/2 , 𝑛o/6 and 𝑛o/8 , respectively. The angle of incidence θ 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 sinθc= (𝑛o/8)/ (𝑛o/6) = 3/4 Now applying Snell’s law in region I and region III 𝑛o sinθ = 𝑛o/6 sinθc Or sinθ = 1/6 sinθc = 1/6 (3/4 ) = 1/8 ∴ θ = sin-1( 1/8 ) .

Q6.  A rectangular glass slab 𝐴𝐵𝐶𝐷, of refractive index 𝑛1, is immersed in water of refractive index 𝑛2(𝑛1 > 𝑛2). A ray of light in incident at the surface 𝐴𝐵 of the slab as shown. The maximum value of the angle of incidence 𝛼𝑚𝑎𝑥 such that the ray comes out only from the other surface 𝐶𝐷 is given by
•   sin-1 [ (𝑛1/𝑛2) cos(sin-1 𝑛2/𝑛1 )]
•   sin-1 [𝑛1 cos(sin-1 1/𝑛2 )]
• sin-1 (𝑛1/𝑛2)
•   sin-1 ( 𝑛2/𝑛1 )
Solution
Ray comes out from 𝐶𝐷, means rays after refraction from 𝐴𝐵 get, total intensity reflected at 𝐴𝐷
– 𝑛1/𝑛2 = sin𝛼max/sin𝑟1 ⇒ 𝛼max = sin-1[ 𝑛1/𝑛2 sin𝑟1] …(i)
Also 𝑟1 + 𝑟2 = 90°
⇒ 𝑟1 = 90 − 𝑟2 = 90 − 𝐶 ⇒ 𝑟1 = 90 − sin-1 ( 1/2𝜇1) ⇒ 𝑟1
= 90 − sin-1 (𝑛2/𝑛1) …(ii)
Hence from equation (i) and (ii)
𝛼max= sin-1 [ 𝑛1/𝑛2 sin(90 − sin-1 𝑛2/𝑛1 )]
= sin-1[(𝑛1/𝑛2)cos(sin-1 (𝑛2/𝑛1)]

Q7.A vessel of depth 2d 𝑐𝑚 is half filled with a liquid of refractive index 𝜇1 and the upper half with a liquid of refractive index 𝜇2. The apparent depth of the vessel seen perpendicularly is
•   𝑑 ( (𝜇1𝜇2)/(𝜇1+ 𝜇2) )
•   𝑑 ( 1/𝜇1 + 1/𝜇2 )
•   2𝑑 ( 1/𝜇1 + 1/𝜇2 )
•   2𝑑 ( 1/𝜇1𝜇2 )
Solution

ℎ′ = 𝑑1/𝜇1 + 𝑑2/𝜇2 = 𝑑 ( 1/𝜇1 + 1/𝜇2 )

Q8. 1% of light of a source with luminous intensity 50 𝑐𝑎𝑛𝑑𝑒𝑙𝑎 is incident on a circular surface of radius 10 𝑐𝑚. The average illuminance of surface is
•   100 𝑙𝑢𝑥
•   200 𝑙𝑢𝑥
•   300 𝑙𝑢𝑥
•   400 𝑙𝑢𝑥
Solution
𝜙= 4𝜋𝐿 = 200 𝜋 𝑙𝑢𝑚𝑒𝑛 So 𝐼 = 𝜙/100 𝐴 = 200 𝜋/100×𝜋𝑟2 = 2/(0.1)2 = 200 𝑙𝑢𝑥

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 𝑚 = 𝑓𝑜/𝑓𝑒

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 𝑖 = 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° ## Want to know more

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BEST NEET COACHING CENTER | BEST IIT JEE COACHING INSTITUTE | BEST NEET, IIT JEE COACHING INSTITUTE: RAY OPTICS AND OPTICAL INSTRUMENTS QUIZ 16
RAY OPTICS AND OPTICAL INSTRUMENTS QUIZ 16
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BEST NEET COACHING CENTER | BEST IIT JEE COACHING INSTITUTE | BEST NEET, IIT JEE COACHING INSTITUTE
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