Ray Optics and Optical Instruments Quiz-14
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.
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Q1. A thin glass (refractive index 1.5) lens has optical power of -5 D in air. Its optical power in a liquid medium with refractive index 1.6 will be
Solution
1/𝑓𝑎 = (𝜇 − 1)(1/𝑅1 −1 /𝑅2) = (1.5 − 1)(1/𝑅1 −1/𝑅2) …………..(𝑖)
1/𝑓𝑎 = (𝜇 − 1)(1/𝑅1 −1 /𝑅2) = (1.5 − 1)(1/𝑅1 −1/𝑅2) …………..(𝑖)
𝑎𝑛𝑑
1/𝑓𝑚 =𝜇𝑔 − 𝜇𝑚/𝜇𝑚(1/𝑅1−1/𝑅2 )
1/𝑓𝑚 = (1.5/1.6− 1)(1/𝑅1- 1/𝑅2) ………..(𝑖𝑖)
Thus,
𝑓𝑚/𝑓𝑎 = (1.5 − 1)/(1.5 1.6 − 1)= −8
𝑓𝑚 = −8 × 𝑓ₐ
= −8 ×−1/5 (∵ 𝑓𝑎 =1/𝑝= −1/5𝑚)
= 1.6 𝑚
∴ 𝑃𝑚 = 𝜇/𝑓𝑚
=1.6/1.6
= 1𝐷
Q2. A bi-convex lens made of glass (refractive index 1.5) is put in a liquid of refractive index 1.7. Its focal length wil
Solution
𝑓 𝑙/𝑓𝑎 = 𝑎𝜇𝑔 − 1/𝑙𝜇𝑔 − 1 = (1.5 − 1) × 1.7/(1.5 − 1.7) ⇒ 𝑓1 = 0.85/−0.2 𝑓𝑎 = −4.25 𝑓𝑎
𝑓 𝑙/𝑓𝑎 = 𝑎𝜇𝑔 − 1/𝑙𝜇𝑔 − 1 = (1.5 − 1) × 1.7/(1.5 − 1.7) ⇒ 𝑓1 = 0.85/−0.2 𝑓𝑎 = −4.25 𝑓𝑎
Q3. If luminous efficiency of a lamp is 2 lumen/watt and its luminous intensity is 42 candela, then power of the lamp is
Solution
Luminous flux = 4𝜋 𝐿 = 4 × 3.14 × 42 = 528 𝐿𝑢𝑚𝑒𝑛 Power of lamp = Luminous flux /Luminous efficiency = 528 /2 = 264 𝑊
Luminous flux = 4𝜋 𝐿 = 4 × 3.14 × 42 = 528 𝐿𝑢𝑚𝑒𝑛 Power of lamp = Luminous flux /Luminous efficiency = 528 /2 = 264 𝑊
Q4. What cause chromatic aberration?
Solution
In chromatic aberration the image formed by a lens has coloured fringes, because the refractive index for different colours is different and hence the focal length of lens for different colours is different. So, the cause of chromatic aberration is the variation of focal length with colour
In chromatic aberration the image formed by a lens has coloured fringes, because the refractive index for different colours is different and hence the focal length of lens for different colours is different. So, the cause of chromatic aberration is the variation of focal length with colour
Q5. Two plane mirrors inclined to each other at an angle 72°, what is the number of image formed
Solution
Number of images formed = 360°/θ = 360°/72 = 5 .
Number of images formed = 360°/θ = 360°/72 = 5 .
Q6. . In an optics experiments, with the position of the object fixed, a student varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance 𝑢 and the image distance 𝑣, from the lens, is plotted using the same scale for the two axes. A straight line passing through the origin and making an angle of 45° with the 𝑥-axis meets the experimental curve at 𝑃. The coordinates of 𝑃will be
Solution
It is possible when object kept at center of curvature
It is possible when object kept at center of curvature
Q7. The aperture of a telescope is made large, because
Solution
For greater aperture of lens light passing through lens is more and so intensity of image increases
For greater aperture of lens light passing through lens is more and so intensity of image increases
Q8. . Four convergent lenses have focal lengths 100 𝑐𝑚,10 𝑐𝑚,4 𝑐𝑚 and 0.3 𝑐𝑚. For a telescope with maximum possible magnification, we choose the lenses of focal length
Solution
𝑚 = − 𝑓𝑜/𝑓 𝑒
𝑚 = − 𝑓𝑜/𝑓 𝑒
Q9. The focal length of a convex lens depends upon
Solution
𝑓 ∝ 1/𝜇−1 and 𝜇 ∝1/𝜆
𝑓 ∝ 1/𝜇−1 and 𝜇 ∝1/𝜆
Q10. A virtual image three times the size of the object is obtained with a concave mirror of curvature 36 𝑐𝑚. The distance of the object from the mirror is
Solution
Image is virtual so 𝑚 = +3 and 𝑓 = 𝑅/2 = 18 𝑐𝑚 So from 𝑚 = 𝑓/𝑓−𝑢 ⇒ 3 = (−18)/(−18)−𝑢 ⇒ 𝑢 = −12 𝑐𝑚
Image is virtual so 𝑚 = +3 and 𝑓 = 𝑅/2 = 18 𝑐𝑚 So from 𝑚 = 𝑓/𝑓−𝑢 ⇒ 3 = (−18)/(−18)−𝑢 ⇒ 𝑢 = −12 𝑐𝑚
