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. .

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
  •   1 D
  •   -1 D
  •   25 D
  • -25 D
1/π‘“π‘Ž = (πœ‡ − 1)(1/𝑅1 −1 /𝑅2) = (1.5 − 1)(1/𝑅1 −1/𝑅2) …………..(𝑖)
 π‘Žπ‘›π‘‘ 1/π‘“π‘š =πœ‡π‘” − πœ‡π‘š/πœ‡π‘š(1/𝑅1−1/𝑅2 )
 1/π‘“π‘š = (1.5/1.6− 1)(1/𝑅1- 1/𝑅2) ………..(𝑖𝑖) 
 π‘“π‘š/π‘“π‘Ž = (1.5 − 1)/(1.5 1.6 − 1)= −8 
 π‘“π‘š = −8 × π‘“β‚ 
 = −8 ×−1/5 (∵ π‘“π‘Ž =1/𝑝= −1/5π‘š)
 = 1.6 π‘š 
∴ π‘ƒπ‘š = πœ‡/π‘“π‘š
= 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
  •   Decrease and change sign
  •   Increase and change sign
  •   Decrease and remain of the same sign
  •   Increase and remain of the same sign
𝑓 𝑙/π‘“π‘Ž = π‘Žπœ‡π‘” − 1/π‘™πœ‡π‘” − 1 = (1.5 − 1) × 1.7/(1.5 − 1.7) ⇒ 𝑓1 = 0.85/−0.2 π‘“π‘Ž = −4.25 π‘“π‘Ž

Q3.   If luminous efficiency of a lamp is 2 lumen/watt and its luminous intensity is 42 candela, then power of the lamp is
  •    62 π‘Š
  •   76 π‘Š
  •   1.38 π‘Š
  •   264 π‘Š
Luminous flux = 4πœ‹ 𝐿 = 4 × 3.14 × 42 = 528 πΏπ‘’π‘šπ‘’π‘› Power of lamp = Luminous flux /Luminous efficiency = 528 /2 = 264 π‘Š

Q4. What cause chromatic aberration?
  •   Non-paraxial rays
  •   Paraxial rays
  •   Variation of focal length with colour
  •   Difference in radii of curvature of the bounding surface of the lens
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
  •   3
  •   5
  •   9
  •   7
 Number of images formed = 360°/ΞΈ = 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 𝑒 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 π‘₯-axis meets the experimental curve at 𝑃. The coordinates of 𝑃will be
  •   (2𝑓,2𝑓)
  •   (𝑓/2 , 𝑓/2 )
  • (𝑓,𝑓)
  •   (4𝑓,4𝑓)
It is possible when object kept at center of curvature
       𝑒 = 𝑣 𝑒 = 2𝑓,𝑣 = 2𝑓.

Q7. The aperture of a telescope is made large, because
  •   To increase the intensity of image
  •   To decrease the intensity of image
  •   To have greater magnification
  •   To have lesser resolution

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 π‘π‘š,10 π‘π‘š,4 π‘π‘š and 0.3 π‘π‘š. For a telescope with maximum possible magnification, we choose the lenses of focal length
  •   100 π‘π‘š,0.3 π‘π‘š
  • 10 π‘π‘š,0.3 π‘π‘š
  •   10 π‘π‘š,4 π‘π‘š
  •   100 π‘π‘š,4 π‘π‘š
π‘š = − π‘“π‘œ/𝑓 𝑒

Q9. The focal length of a convex lens depends upon

  •   Frequency of the light ray
  •   Wavelength of the light ray
  •   Both (a) and (b)
  •   None of these
𝑓 ∝ 1/πœ‡−1 and πœ‡ ∝1/πœ†

Q10.  A virtual image three times the size of the object is obtained with a concave mirror of curvature 36 π‘π‘š. The distance of the object from the mirror is
  •   5 π‘π‘š
  •   12 π‘π‘š
  •   10 π‘π‘š
  • 20 π‘π‘š
Image is virtual so π‘š = +3 and 𝑓 = 𝑅/2 = 18 π‘π‘š So from π‘š = 𝑓/𝑓−𝑒 ⇒ 3 = (−18)/(−18)−𝑒 ⇒ 𝑒 = −12 π‘π‘š

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