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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.  A double convex lens of focal length 20 𝑐𝑚 is made of glass of refractive index 3/2. When placed completely in water ( 𝑎𝜇𝜔 = 4/3), its focal length will be
•   80 𝑐𝑚
•   15 𝑐𝑚
•   17.7 𝑐𝑚
•   22.5 𝑐𝑚
Solution
𝑓water = 4 × 𝑓air , air lens is made up of glass

Q2. To a fish under water, viewing obliquely a fisherman standing on the bank of the lake, the man looks
•   Taller than what he actually is
•   shorter that what he actual is
•   The same height as he actually is
•   Depends on the obliquity
Solution
When a ray from main 𝑂 in air (rarer medium) goes to water (denser medium), then it bends towards the normal. Extent 𝑀𝑁 backwards meet at point 𝑂′. Therefore, it appears to the fish as if the man is taller than what he actually is

Q3.   The numerical aperture for a human eye is of the order of
•    1
•   0.1
•   0.01
•   0.001
Solution
0.001

Q4.  Five 𝑙𝑢𝑚𝑒𝑛/𝑤𝑎𝑡𝑡 is the luminous efficiency of a lamp and its luminous intensity is 35 𝑐𝑎𝑛𝑑𝑒𝑙𝑎. The power of the lamp is
•   80 𝑊
•   176 𝑊
•   88 𝑊
•   36 𝑊
Solution
Efficiency of light source 𝜂 = 𝜙/𝑃 …(i) and 𝐿 = 𝜙/4𝜋 …(ii) From equation (i) and (ii) ⇒ 𝑃 = 4𝜋𝐿/𝜂 = (4𝜋 × 35) /5 ≈ 88 𝑊

Q5. If the speed of light in vacuum is 𝐶 𝑚/𝑠𝑒𝑐, then the velocity of light in a medium of refractive index 1.5
•   Is 1.5 × 𝐶
•   Is 𝐶
•   Is 𝐶/1.5
•   can have any velocity
Solution
𝜇 = 𝐶/𝐶𝑚 ⇒ 𝐶𝑚 = 𝐶/1.5 .

Q6.  When sunlight is scattered by minute particles of atmosphere, the intensity of light scattered away is proportional to
•   (wavelength of light)4
•   (frequency of light)4
• (wavelength of light)2
•   (frequency of light)2
Solution
(frequency of light)4

Q7. A ray of light is incident at the glass-water interface at an angle 𝑖it emerges finally parallel to the surface of water, then the value of μgwould be
•   (4/3) 𝑠𝑖𝑛𝑖
•   1/𝑠𝑖𝑛𝑖
•   4/3
•   1
Solution

For glass-water interface gμw = sin𝑖/sin𝑟
For water air interface wμ𝑎 = sin𝑟/sin90°
⇒ gμw × wμ𝑎 = sin𝑖/sin𝑟 × sin𝑟/sin90° = sin𝑖 or μwg × μ𝑎w = sin𝑖
⇒ μg= 1/sin𝑖

Q8. . The wavelength of light in air and some other medium are respectively 𝜆𝑎 and 𝜆𝑚. The refractive index of medium is
•   𝜆𝑎/𝜆𝑚
• 𝜆𝑚/𝜆𝑎
•   𝜆𝑎 × 𝜆𝑚
•   None of these
Solution
𝜇𝑚 = 𝑐/𝑣 = 𝑛𝜆𝑎/𝑛𝜆𝑚 = 𝜆𝑎/𝜆𝑚

Q9. The plane face of a planoconvex lens is silvered. If μ be the refractive index and 𝑅, the radius of curvature of curved surface, then the system will behave like a concave mirror of radius of curvature

•   μ 𝑅
•   𝑅/(μ − 1 )
•   R2/ μ
•   [ (μ + 1)/(μ − 1) ]𝑅
Solution
When an object is placed in front of such a lens, the rays are first of all refracted from the convex surface and again refracted from convex surface. Let 𝑓₁,𝑓m be focal lengths of convex surface and mirror (plane polished surface) respectively, then effective focal length is
1 𝐹 = 1/𝑓₁ + 1/𝑓𝑚 + 1/𝑓₁ = 2/𝑓₁ + 1/𝑓𝑚
Since, 𝑓𝑚 =𝑅/2= ∞ ∴1/𝐹=2/𝑓₁
From lens formula 1/𝑓₁ = (𝜇 − 1)( 1 𝑅 )
∴1/𝐹=2(𝜇 − 1)/ 𝑅 ⟹ 𝐹 = 𝑅/2(𝜇 − 1)
𝑜𝑟 𝑅𝑒𝑞 = 2𝐹 =𝑅 /(𝜇 − 1)

Q10.  The slab of a material of refractive index 2 shown in figure has curved surface 𝐴𝑃𝐵 of radius of curvature 10 𝑐𝑚 and a plane surface 𝐶𝐷. On the left of 𝐴𝑃𝐵 is air and on the right of 𝐶𝐷 is water with refractive indices as given in figure. An object 𝑂 is placed at a distance of 15 𝑐𝑚 from pole 𝑃 as shown. The distance of the final image of 𝑂 from 𝑃, as viewed from the left is
•   20cm
•   30cm
•   40cm
•   50cm
Solution
In case of refraction from a curved surface, we have
𝜇2/𝑣 − 𝜇1/𝑢 = (𝜇2 − 𝜇1)/ 𝑅 ⇒ 1/𝑣 − 2 /(−15) = (1 − 2) /−10 ⇒ 𝑣 = −30 𝑐𝑚
𝑖.𝑒., the curved surface will form virtual image 𝐼 at distance of 30 𝑐𝑚 from 𝑃. Since the image is virtual there will be no refraction at the plane surface 𝐶𝐷 (as the rays are not actually passing through the boundary), the distance of final image 𝐼 from 𝑃 will remain 30 𝑐𝑚 ## Want to know more

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BEST NEET COACHING CENTER | BEST IIT JEE COACHING INSTITUTE | BEST NEET & IIT JEE COACHING: RAY OPTICS AND OPTICAL INSTRUMENTS QUIZ 20
RAY OPTICS AND OPTICAL INSTRUMENTS QUIZ 20
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