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.**A monochromatic beam of light passes from a denser medium into a rarer medium. As a result

Solution

𝑣 ∝ 1/𝜇 ,𝜇rarer < 𝜇denser

𝑣 ∝ 1/𝜇 ,𝜇rarer < 𝜇denser

**Q2.**In an astronomical telescope in normal adjustment, a straight black line of length 𝐿 is drawn on the objective lens. The eyepiece forms a real image of this line. The length of this image is 𝑙. The magnification of the telescope is

Solution

Here we treat the line on the objective as the object and the eyepiece as the lens

Here we treat the line on the objective as the object and the eyepiece as the lens

Hence 𝑢 = −(𝑓

_{o}+ 𝑓_{e}) and 𝑓 = 𝑓_{e}Now 1/𝑣 − 1 /−(𝑓

_{o}+𝑓_{e}) = 1/𝑓_{e}Solving we get 𝑣 = (𝑓

_{o}+𝑓_{e})𝑓_{e}/𝑓_{o}Magnification = |𝑣/𝑢| = 𝑓

_{e}/𝑓_{o}= Image size/Object size = 𝑙/𝐿∴ Magnification of telescope in normal adjustment
=
𝑓

_{0}/𝑓_{e}= 𝐿/𝑙**Q3.**A microscope is focused on an ink mark on the top of a table. If we place a glass slab of 3 cm thick on it, how should the microscope be moved to focus the ink spot again? The refractive index of glass is 1.5.

Solution

Given, 𝑛 = 1.5, ℎ

Given, 𝑛 = 1.5, ℎ

_{0}= 3 𝑛 =
ℎ

_{0}ℎ_{1} 1.5 =
3/ℎ

_{1} ℎ

_{i}= 3/1.5 = 2 𝑐𝑚 Hence, 3-2 = 1 cm upwards

**Q4.**With respect to air critical angle in a medium for light of red colour (𝜆1) is 𝜃. Other facts remaining same, critical angle for light of yellow colour [𝜆2] will be

Solution

Critical angle = sin

Critical angle = sin

^{-1}(1/𝜇)∴ 𝜃 = sin

^{-1}( 1/𝜇_{𝜆1}) and 𝜃′ = sin^{-1}( 1 𝜇_{𝜆2}) Since 𝜇

_{𝜆2}> 𝜇_{𝜆1}, hence 𝜃′ < 𝜃**Q5.**Total internal reflection of a ray of light is possible when the (𝑖

_{c}= critical angle, 𝑖 = angle of incidence)

Solution

**Ray goes from denser medium to rarer medium and 𝑖 > 𝑖**._{c}**Q6.**What is the time taken (in 𝑠𝑒𝑐𝑜𝑛𝑑𝑠) to cross a glass of thickness 4 𝑚𝑚 and 𝜇 = 3 by light

Solution

𝑡 = 𝜇𝑥/𝑐 = (3 × 4 × 10

𝑡 = 𝜇𝑥/𝑐 = (3 × 4 × 10

^{-3)}/(3 × 10^{8}) = 4 × 10^{-11}𝑠**Q7.**. A compound microscope has an objective and eye-piece as thin lenses of focal lengths 1 cm and 5 cm respectively. The distance between the objective and the eye-piece is 20 cm. The distance at which the object must be placed infront of the objective if the final image is located at 25 cm from the eye-piece, it numerically

Solution

For the eye-piece 𝑣

For the eye-piece 𝑣

_{e}= −25 cm,𝑓_{e}= 5 cm 1/−25 − 1/𝑢

_{e}= 1/5Or
1/𝑢

_{e}= − 1/25 − 1/5or
1/𝑢

_{e}= − −1−5/25 Or 𝑢

_{e}= − 25/6Now , 𝑣

_{o}= 𝐿 − |𝑢_{e}| = 20 − 25/6= (120 − 25)/6cm =95/6cm Now, 1 /(95/6)−1/𝑢

_{o}= 1/1 or 1/𝑢_{O}=6/95− 1 Or 𝑢

_{0}= −95/89cm or |𝑢_{o}| = 95/89cm**Q8.**From which source a continuous emission spectrum and a line absorption spectrum are simultaneously obtained

Solution

The sun

The sun

**Q9.**In a thin prism of glass (refractive index 1.5), which of the following relations between the angle of minimum deviations 𝛿𝑚 and angle of refraction 𝑟 will be correct

Solution

𝛿𝑚 = (𝜇 − 1)(2𝑟) = (1.5 − 1)2𝑟 = 0.5 × 2𝑟 = 𝑟

𝛿𝑚 = (𝜇 − 1)(2𝑟) = (1.5 − 1)2𝑟 = 0.5 × 2𝑟 = 𝑟

**Q10.**A person cannot see properly beyond 2 m. Power of the lens is

Solution

Focal length of eye lens = −2 𝑚 So , power of lens = 1/𝑓 = 1/−2 = -0.5 D (short-sighted)

Focal length of eye lens = −2 𝑚 So , power of lens = 1/𝑓 = 1/−2 = -0.5 D (short-sighted)