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.
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Q1. The index of refraction of diamond is 2.0. The velocity of light in diamond is approximately
Solution
Refractive index of diamond is μ = 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
Refractive index of diamond is μ = 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
Solution
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.
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
Solution
Due to the absorption of certain wavelengths by the elements in outer layers of sun
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?
Solution
𝑚 = 𝑓/𝑓 − 𝑢 1/2 = 200/(200 − 𝑢) 200 – 𝑢 = 400 𝑢 = −200 cm 𝑢 = −2 m
𝑚 = 𝑓/𝑓 − 𝑢 1/2 = 200/(200 − 𝑢) 200 – 𝑢 = 400 𝑢 = −200 cm 𝑢 = −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
Solution
𝛿 ∝ (𝜇 − 1) .
𝛿 ∝ (𝜇 − 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]
Solution
For first lens, 𝜇1 = −30𝑐𝑚,𝑓1 = 10𝑐𝑚
For first lens, 𝜇1 = −30𝑐𝑚,𝑓1 = 10𝑐𝑚
1/𝑓 = 1/𝑣 − 1/𝑢
𝑜𝑟 1/𝑣=1/𝑓+1/𝑢
𝑜𝑟 1/𝑣=1/10−1/30=1/15
𝑜𝑟 𝑣 = 15 𝑐𝑚
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 , 𝜇2 =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 𝑐𝑚
Q7. The focal length of a convex lens is 10 𝑐𝑚 and its refractive index is 1.5. If the radius of curvature of one surface is 7.5 𝑐𝑚, the radius of curvature of the second surface will be
Solution
1/𝑓 = (𝜇 − 1)( 1/ 𝑅1 − 1/𝑅2 ) ⇒ 1 /+10 = (1.5 − 1)( 1/+7.5 − 1/𝑅2 ) ⇒ 𝑅2 = −15 𝑐𝑚
1/𝑓 = (𝜇 − 1)( 1/ 𝑅1 − 1/𝑅2 ) ⇒ 1 /+10 = (1.5 − 1)( 1/+7.5 − 1/𝑅2 ) ⇒ 𝑅2 = −15 𝑐𝑚
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 𝑥. If numerical value of magnification produced by lens is 𝑚, focal length of lens is
Solution
Here, 𝑥 = 𝑢 + 𝑣 As 𝑚 = 𝑓/(𝑓+𝑢) = (𝑓−𝑣)/𝑓 and image is real, magnification is negative ∴ −𝑚 = 𝑓/(𝑓 + 𝑢),𝑢 = −(𝑚 + 1)𝑓/ 𝑚 From –𝑚 = (𝑓−𝑣) /𝑓 ⇒ 𝑣 = (𝑚 + 1)𝑓 Put in Eq.(i) 𝑥 =−((𝑚 + 1)/ 𝑚)𝑓 + (𝑚 + 1)𝑓 Solving ,we get, 𝑓 = 𝑚𝑥/(𝑚+1)2
Here, 𝑥 = 𝑢 + 𝑣 As 𝑚 = 𝑓/(𝑓+𝑢) = (𝑓−𝑣)/𝑓 and image is real, magnification is negative ∴ −𝑚 = 𝑓/(𝑓 + 𝑢),𝑢 = −(𝑚 + 1)𝑓/ 𝑚 From –𝑚 = (𝑓−𝑣) /𝑓 ⇒ 𝑣 = (𝑚 + 1)𝑓 Put in Eq.(i) 𝑥 =−((𝑚 + 1)/ 𝑚)𝑓 + (𝑚 + 1)𝑓 Solving ,we get, 𝑓 = 𝑚𝑥/(𝑚+1)2
Q9. An object of height 1.5 𝑐𝑚 is placed on the axis of a convex lens of focal length 25 𝑐𝑚. A real image is formed at a distance of 75 𝑐𝑚 from the lens. The size of the image will be
Solution
𝐼/𝑂 = (𝑓 − 𝑣)/𝑓 ⇒𝐼 /+1.5 =(25 − 75) /25= −2 ⇒ 𝐼 = −3 𝑐𝑚
𝐼/𝑂 = (𝑓 − 𝑣)/𝑓 ⇒𝐼 /+1.5 =(25 − 75) /25= −2 ⇒ 𝐼 = −3 𝑐𝑚
Q10. An achromatic combination of lenses is formed by joining
Solution
For an achromatic combination 𝜔1/𝑓1 + 𝜔2/𝑓2 = 0 𝑖.𝑒., 1 convex lens and 1 concave lens
For an achromatic combination 𝜔1/𝑓1 + 𝜔2/𝑓2 = 0 𝑖.𝑒., 1 convex lens and 1 concave lens
