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.
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**Q1.**The unit of focal power of a lens is

Solution

Dioptre

Dioptre

**Q2.**Two lenses of power – 15D and +5 D are in contact with each other. The focal length of the combination is

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

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

Solution

Minimum angular separation ∆𝜃 = 1/𝑅.𝑃. = 1.22 𝜆/𝑑 = (1.22 × 5000 × 10−10)/ 2 = 0.3 × 10

Minimum angular separation ∆𝜃 = 1/𝑅.𝑃. = 1.22 𝜆/𝑑 = (1.22 × 5000 × 10−10)/ 2 = 0.3 × 10

^{-6}𝑟𝑎𝑑

**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

Solution

As refractive index for 𝑧 > 0 and 𝑧 ≤ 0 is different 𝑋 − 𝑌 plane should be boundary between two media Angle of incidence,

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/
√𝐴

_{x}^{2}+ 𝐴_{y}^{2}+ 𝐴_{z}^{2}| | = 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 𝑛

_{o}, 𝑛

_{o}/2 , 𝑛

_{o}/6 and 𝑛

_{o}/8 , respectively. The angle of incidence θ for which the beam just misses entering Region IV is

Solution

**Critical angle from region III to region IV sinθ**._{c}= (𝑛_{o}/8)/ (𝑛_{o}/6) = 3/4 Now applying Snell’s law in region I and region III 𝑛_{o}sinθ = 𝑛_{o}/6 sinθ_{c}Or sinθ = 1/6 sinθ_{c}= 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 𝑛

_{2}(𝑛

_{1}> 𝑛

_{2}). 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

Solution

Ray comes out from 𝐶𝐷, means rays after refraction from 𝐴𝐵 get, total intensity reflected at 𝐴𝐷

Ray comes out from 𝐶𝐷, means rays after refraction from 𝐴𝐵 get, total intensity reflected at 𝐴𝐷

– 𝑛

_{1}/𝑛_{2}= sin𝛼_{max}/sin𝑟_{1}⇒ 𝛼_{max}= sin^{-1}[ 𝑛_{1}/𝑛_{2}sin𝑟_{1}] …(i) Also 𝑟

_{1}+ 𝑟_{2}= 90°⇒ 𝑟

_{1}= 90 − 𝑟_{2}= 90 − 𝐶 ⇒ 𝑟_{1}= 90 − sin^{-1}( 1/2𝜇_{1}) ⇒ 𝑟_{1}= 90 − sin

^{-1}(𝑛_{2}/𝑛_{1}) …(ii)Hence from equation (i) and (ii)

𝛼

_{max}= sin^{-1}[ 𝑛_{1}/𝑛_{2}sin(90 − sin^{-1}𝑛_{2}/𝑛_{1})]= sin

^{-1}[(𝑛_{1}/𝑛_{2})cos(sin^{-1}(𝑛_{2}/𝑛_{1})]**Q7.**A vessel of depth 2d 𝑐𝑚 is half filled with a liquid of refractive index 𝜇

_{1}and the upper half with a liquid of refractive index 𝜇

_{2}. The apparent depth of the vessel seen perpendicularly is

Solution

ℎ′ = 𝑑

ℎ′ = 𝑑

_{1}/𝜇_{1}+ 𝑑_{2}/𝜇_{2}= 𝑑 ( 1/𝜇_{1}+ 1/𝜇_{2})**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

Solution

𝜙= 4𝜋𝐿 = 200 𝜋 𝑙𝑢𝑚𝑒𝑛 So 𝐼 = 𝜙/100 𝐴 = 200 𝜋/100×𝜋𝑟2 = 2/(0.1)

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

Solution

In case of a telescope if object and final image are at infinity then 𝑚 = 𝑓𝑜/𝑓𝑒

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

Solution

From the figure it is clear that the angle between incident ray and emergent ray is 90°

From the figure it is clear that the angle between incident ray and emergent ray is 90°