Ray Optics and Optical Instruments Quiz-17

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.  The index of refraction of diamond is 2.0. The velocity of light in diamond is approximately
  •   1.5 × 1010cms-1
  •   2 × 1010cms-1
  •   3.0 × 1010cms-1
  •   6 ×1010cms-1
Refractive index of diamond is ΞΌ = velocity of light in air/velocity of light in diamond 2 = 3.0 × 1010/ velocity of light in diamond So, velocity of light in diamond is = (3.0 × 1010)/2 = 1.5 ×1010cms-1

Q2. Image formed by a convex lens is virtual and erect when the object is placed
  •   At 𝐹
  •   Between 𝐹 and the lens
  •   At 2𝐹
  •   Beyond 2𝐹
When object is placed between 𝐹 and pole of a convex lens then a virtual, erect and magnified image will be formed on the same side behind the object.

Q3.   Dark lines on solar spectrum are due to
  •   Lack of certain elements
  •   Black body radiation
  •   Absorption of certain wavelengths by outer layers
  •   Scattering
Due to the absorption of certain wavelengths by the elements in outer layers of sun

Q4.  Where should a person stand straight from the pole of a convex mirror of focal length 2.0 m on its axis so that the image formed become half of his original height?
  •   -2.60m
  •   -4.0m
  •   -0.5m
  •   -2.0m
π‘š = 𝑓/𝑓 − 𝑒 1/2 = 200/(200 − 𝑒) 200 – 𝑒 = 400 𝑒 = −200 cm 𝑒 = −2 m

Q5.Three prisms 1, 2 and 3 have the prism angle 𝐴 = 60°, but their refractive indices are respectively 1.4, 1.5 and 1.6. If 𝛿1,𝛿2,𝛿34 be their respective angles of deviation then
  •   𝛿3 > 𝛿2 > 𝛿1
  •   𝛿1 > 𝛿2 > 𝛿3
  •   𝛿1 = 𝛿2 = 𝛿3
  •   𝛿2 > 𝛿1 > 𝛿3
  𝛿 ∝ (πœ‡ − 1) .

Q6.  . The position of final image formed by the given lens combination from the third lens will be at a distance of [𝑓1 = +10 cm,𝑓2= −10 cm,𝑓3 = +30 cm] 

  •   15 cm
  •   Infinity
  • sin-1 45 cm
  •   30 cm
For first lens, πœ‡1 = −30π‘π‘š,𝑓1 = 10π‘π‘š
 1/𝑓 = 1/𝑣 − 1/𝑒
 π‘œπ‘Ÿ 1/𝑣=1/𝑓+1/𝑒 
π‘œπ‘Ÿ 1/𝑣=1/10−1/30=1/15
 π‘œπ‘Ÿ 𝑣 = 15 π‘π‘š Therefore, image formed by convex lens (𝐿1) is at point 𝐼1 and acts as virtual object for concave lens 

The image 𝐼1 is formed at focus of concave lens (as shown) and so emergent rays will be parallel to the principle axis. For lens 𝐿2 , πœ‡2 =15-5=10 cm, 𝑓2=-10cm. These parallel rays are incident on the third convex lens (𝐿3) and will be brought to convergence at the focus of the lens (𝐿3) Hence , distance of final image from third lens 𝐿3 𝑣2= 𝑓3 = 30 π‘π‘š

Q7. The focal length of a convex lens is 10 π‘π‘š and its refractive index is 1.5. If the radius of curvature of one surface is 7.5 π‘π‘š, the radius of curvature of the second surface will be
  •   7.5 π‘π‘š
  •   15.0 π‘π‘š
  •   75 π‘π‘š
  •   5.0 π‘π‘š

1/𝑓 = (πœ‡ − 1)( 1/ 𝑅1 − 1/𝑅2 ) ⇒ 1 /+10 = (1.5 − 1)( 1/+7.5 − 1/𝑅2 ) ⇒ 𝑅2 = −15 π‘π‘š

Q8. A convex lens of focal length 𝑓 is placed some where in between an object and a screen. The distance between object and screen is π‘₯. If numerical value of magnification produced by lens is π‘š, focal length of lens is
  •   π‘šπ‘₯/(π‘š + 1)2
  •   π‘šπ‘₯/(π‘š − 1)2
  •   (π‘š + 1)2/ π‘š π‘₯
  •   (π‘š − 1)2/π‘š π‘₯
Here, π‘₯ = 𝑒 + 𝑣 As π‘š = 𝑓/(𝑓+𝑒) = (𝑓−𝑣)/𝑓 and image is real, magnification is negative ∴ −π‘š = 𝑓/(𝑓 + 𝑒),𝑒 = −(π‘š + 1)𝑓/ π‘š From –π‘š = (𝑓−𝑣) /𝑓 ⇒ 𝑣 = (π‘š + 1)𝑓 Put in Eq.(i) π‘₯ =−((π‘š + 1)/ π‘š)𝑓 + (π‘š + 1)𝑓 Solving ,we get, 𝑓 = π‘šπ‘₯/(π‘š+1)2

Q9. An object of height 1.5 π‘π‘š is placed on the axis of a convex lens of focal length 25 π‘π‘š. A real image is formed at a distance of 75 π‘π‘š from the lens. The size of the image will be

  •   4.5 π‘π‘š
  •   3.0 π‘π‘š
  •   0.75 π‘π‘š
  •   0.5 π‘π‘š
𝐼/𝑂 = (𝑓 − 𝑣)/𝑓 ⇒𝐼 /+1.5 =(25 − 75) /25= −2 ⇒ 𝐼 = −3 π‘π‘š

Q10.  An achromatic combination of lenses is formed by joining
  •   2 convex lenses
  •   2 concave lenses
  •   1 convex lens and 1 concave lens
  •   Convex lens and plane mirror
For an achromatic combination πœ”1/𝑓1 + πœ”2/𝑓2 = 0 𝑖.𝑒., 1 convex lens and 1 concave lens

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