Friday, February 7, 2025

Ray Optics and Optical Instruments

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Ray Optics and Optical Instruments is a crucial chapter in 12th-grade physics (NCERT) that explores how light behaves as rays and how we use this behavior to build optical instruments. Here’s a summary:

I. Ray Optics: The Basics

  • Reflection: Light bouncing off a surface. Key concepts include the laws of reflection (angle of incidence equals angle of reflection), types of mirrors (plane, concave, convex), and the mirror formula (1/f = 1/v + 1/u) relating focal length (f), image distance (v), and object distance (u). Magnification (m = -v/u) tells us about the image size and orientation.
  • Refraction: Light bending when passing from one medium to another. Snell’s Law (n₁sinθ₁ = n₂sinθ₂) governs this, where ‘n’ represents refractive index and ‘θ’ represents the angle. Total Internal Reflection (TIR) occurs when light travels from a denser to a rarer medium at an angle greater than the critical angle, leading to applications like optical fibers.
  • Dispersion: White light splitting into its constituent colors due to different wavelengths having different refractive indices. Prisms are a classic example.

II. Refraction at Spherical Surfaces & Lenses

  • Lenses: Converging (convex) and diverging (concave) lenses form images by refracting light. The Lens-maker’s formula helps determine focal length based on the lens’s shape and refractive index. The lens formula (same as the mirror formula but with a sign difference) is used to analyze image formation. Power of a lens (P = 1/f) is measured in diopters.
  • Image Formation by Lenses: Understanding how lenses form real and virtual images is essential. Ray diagrams are helpful tools.

III. Optical Instruments

This section focuses on how combinations of lenses (and sometimes mirrors) are used to create instruments that extend our vision:

  • The Human Eye: A natural optical instrument with a lens that focuses light onto the retina. Common defects like myopia (nearsightedness) and hypermetropia (farsightedness) are corrected using lenses.
  • Microscopes: Used to magnify tiny objects. Compound microscopes use an objective lens and an eyepiece to achieve high magnification. Magnifying power is a key concept.
  • Telescopes: Used to view distant objects. Astronomical telescopes use lenses or mirrors to gather light and form an image. Angular magnification is crucial here.

EXERCISES

1. A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?

Ans :

2. A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

Ans : 

Therefore, image is virtual, formed at 6.67 cm at the back of the mirror.

3. A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

Ans : 

1. Calculate the refractive index of water:

  • Refractive index (n) 

= Real depth / Apparent depth

  • n = 12.5 cm / 9.4 cm
  • n ≈ 1.33

2. Calculate the new apparent depth with the new liquid:

  • New apparent depth = Real depth / New refractive index
  • New apparent depth = 12.5 cm / 1.63
  • New apparent depth ≈ 7.67 cm

3. Calculate the difference in apparent depths:

  • Difference = Original apparent depth – New apparent depth
  • Difference = 9.4 cm – 7.67 cm
  • Difference ≈ 1.73 cm

Answer:

  • The refractive index of water is approximately 1.33.
  • The microscope would have to be moved approximately 1.73 cm closer to the tank.

4. Figures 9.27(a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45° with the normal to a water-glass interface [Fig. 9.27(c)]

Ans :

(a)

(b)

(c)

5. A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)

Ans : 

1. Find the critical angle:

  • The critical angle is the angle of incidence at which light traveling from a denser medium (water) to a rarer medium (air) is refracted at an angle of 90 degrees.
  • We can use Snell’s Law to find it: n₁sinθ₁ = n₂sinθ₂
    • n₁ = refractive index of water (1.33)
    • θ₁ 

= critical angle (what we want to find)

  • n₂ = refractive index of air (approximately 1)
  • θ₂ = 90 degrees
  • Substituting the values: 1.33 * sin(θc) = 1 * sin(90°)
  • Solving for θc: sin(θc) = 1 / 1.33 => θc ≈ 48.75 degrees

2. Visualize the light cone:

  • Imagine a cone of light originating from the bulb at the bottom of the tank. Only the light rays within this cone (with an angle less than the critical angle from the normal) will be able to emerge from the water.

3. Find the radius of the circle of light on the surface:

  • The radius (r) of this circle is related to the depth (d) of the water by the tangent of the critical angle:
    • tan(θc) = r / d
  • Convert depth to meters: 80 cm = 0.8 meters
  • Calculate the radius: r = d * tan(θc) = 0.8 m * tan(48.75°) ≈ 0.91 meters

4. Calculate the area of the circle:

  • Area (A) = πr² = π * (0.91 m)² ≈ 2.60 m²

The area of the water surface through which light from the bulb can emerge is approximately 2.60 square meters.

6. A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.

Ans : 

Part 1: Refractive Index of the Prism

  • Formula: The refractive index (n) of a prism is related to the angle of minimum deviation (δm) and the prism angle (A) by the following formula:

    n = sin((A + δm) / 2) / sin(A / 2)
  • Given:
    • δm = 40° (angle of minimum deviation)
    • A = 60° (prism angle)
  • Calculation:
    • n = sin((60° + 40°) / 2) / sin(60° / 2)
    • n = sin(50°) / sin(30°)
    • n ≈ 1.53

Part 2: New Angle of Minimum Deviation in Water

When the prism is placed in water, we need to consider the relative refractive index of the glass with respect to water.

  1. Relative Refractive Index:
    • n_relative = n_glass / n_water = 1.53 / 1.33 ≈ 1.15
  2. New Angle of Minimum Deviation:
    • Use the same formula as before, but with the relative refractive index and the new angle of minimum deviation (δm’):

      n_relative = sin((A + δm’) / 2) / sin(A / 2)
    • Substitute the values: 1.15 = sin((60° + δm’) / 2) / sin(30°)
    • Solve for δm’:
      • 1.15 * sin(30°) = sin((60° + δm’) / 2)
      • 0.575 = sin((60° + δm’) / 2)
      • arcsin(0.575) = (60° + δm’) / 2
      • 35.1° ≈ (60° + δm’) / 2
      • 70.2° ≈ 60° + δm’
      • δm’ ≈ 10.2°
  • The refractive index of the glass prism is approximately 1.53.
  • The new angle of minimum deviation when the prism is placed in water is approximately 10.2°.

7. Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?

Ans : 

1. Use the Lens Maker’s Formula:

The Lens Maker’s Formula relates the focal length (f) of a lens to its refractive index (n) and the radii of curvature of its two surfaces (R₁ and R₂):

1/f = 

(n – 1) * (1/R₁ – 1/R₂)

2. Apply the given information:

  • focal length (f): 20 cm
  • refractive index (n): 1.55
  • radii of curvature: The problem states that both faces have the same radius of curvature. Let’s call this R. Since it’s a double-convex lens, both radii are positive. R₁ = R and R₂ = -R (because it’s convex on both sides)

3. Substitute and solve:

1/20 = (1.55 – 1) * (1/R – 1/(-R)) 1/20 = 0.55 * (1/R + 1/R) 1/20 = 0.55 * (2/R) 1/20 = 1.1/R R = 1.1 * 20 R = 22 cm

The radius of curvature required for each face is 22 cm.

8. A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is

(a) a convex lens of focal length 20 cm, and

(b) a concave lens of focal length 16 cm?

Ans : 

(a) Convex Lens (f = 20 cm):

  1. Sign Conventions: Since the beam is converging towards P, the object at P is considered a virtual object. For virtual objects, the object distance (u) is positive. The focal length of a convex lens is also positive.
  2. Lens Formula: Use the lens formula: 1/f = 1/v – 1/u
    • f = focal length = 20 cm
    • u = object distance = 12 cm
    • = image distance (what we want to find)
  3. Solve for v:
    • 1/20 = 1/v – 1/12
    • 1/v = 1/20 + 1/12
    • 1/v = (3 + 5) / 60
    • 1/v = 8/60
    • v = 60/8 = 7.5 cm
  4. Interpretation: The positive value of v means the beam converges 7.5 cm after the lens.

(b) Concave Lens (f = -16 cm):

  1. Sign Conventions: The object is still virtual (u is positive), but the focal length of a concave lens is negative.
  2. Lens Formula: Use the lens formula again: 1/f = 1/v – 1/u
    • f = focal length = -16 cm
    • u = object distance = 12 cm
    • = image distance (what we want to find)
  3. Solve for v:
    • 1/-16 = 1/v – 1/12
    • 1/v = 1/12 – 1/16
    • 1/v = (4 – 3) / 48
    • 1/v = 1/48
    • v = 48 cm

9. An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?

Ans :

10. What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.

Ans : 

1. Use the combined focal length formula:

When two thin lenses are in contact, their effective focal length (F) is given by:

1/F = 1/f₁ + 1/f₂  

Where:

  • f₁ is the focal length of the first lens (convex lens = +30 cm)
  • f₂ is the focal length of the second lens (concave lens = -20 cm)

2. Substitute and solve:

1/F = 1/30 + 1/(-20) 1/F = 1/30 – 1/20 1/F = (2 – 3) / 60 1/F = -1/60 F = -60 cm

11. A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?

Ans : 

(a)

(b)

12. A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.

Ans : 

1. Convert Units:

  • For consistency, let’s work in centimeters (cm):
    • Objective focal length (f₀) = 8.0 mm = 0.8 cm
    • Object distance (u₀) = 9.0 mm = 0.9 cm

2. Find the Image Distance for the Objective (v₀):

  • Use the lens formula for the objective lens:
    • 1/f₀ = 1/v₀ – 1/u₀
    • 1/0.8 = 1/v₀ – 1/0.9
  • Solve for v₀:
    • 1/v₀ = 1/0.8 + 1/0.9
    • v₀ ≈ 7.2 cm

3. Calculate the Separation Between Lenses (L):

  • The separation is the sum of the objective’s image distance and the eyepiece’s focal length:
    • L = v₀ + fₑ
    • L = 7.2 cm + 2.5 cm = 9.7 cm

4. Calculate the Magnification of the Objective (M₀):

  • M₀ = -v₀ / u₀
  • M₀ = -7.2 cm / 0.9 cm = -8

5. Calculate the Magnification of the Eyepiece (Mₑ):

  • Since the eye is focused at the near point (25 cm), we use:
    • Mₑ = 1 + (D / fₑ) where D is the near point distance
    • Mₑ = 1 + (25 cm / 2.5 cm) = 11

6. Calculate the Total Magnification (M):

  • M = M₀ * Mₑ
  • M = 8 * 11 = 88

13. A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?

Ans : 

1. Magnifying Power (M):

The magnifying power of a telescope is given by the formula:

M = -f₀ / fₑ

Where:

  • f₀ = focal length of the objective lens (144 cm)
  • fₑ = focal length of the eyepiece (6.0 cm)

Substituting the values:

M = -144 cm / 6.0 cm = -24  

The magnifying power is 24. The negative sign indicates that the final image is inverted, which is typical for astronomical telescopes.  

2. Separation between Objective and Eyepiece:

For a telescope in normal adjustment (when the final image is formed at infinity), the separation between the objective and eyepiece is simply the sum of their focal lengths:

Separation = f₀ + fₑ

Substituting the values:

Separation = 144 cm + 6.0 cm = 150 cm

14. (a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope?

(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 × 106 m and the radius of lunar orbit is 3.8 × 108 sm.

Ans : 

(a) Angular Magnification:

M = -f₀ / fₑ

Where:

  • f₀ = focal length of the objective lens (15 m = 1500 cm)
  • fₑ = focal length of the eyepiece (1.0 cm)

Substituting the values:

M = -1500 cm / 1.0 cm = -1500

The angular magnification is 1500. 

(b) Diameter of the Moon’s Image:

We can use the concept of similar triangles to relate the diameter of the moon (Dₘ), the diameter of the image (Dᵢ), the focal length of the objective (f₀), and the radius of the lunar orbit (R).

The ratio of the image size to the object size is approximately equal to the ratio of the focal length of the objective to the distance to the object (for distant objects). In this case, the object distance is the radius of the lunar orbit.

Dᵢ / Dₘ ≈ f₀ / R

Where:

  • Dᵢ = diameter of the moon’s image (what we want to find)
  • Dₘ = diameter of the moon (3.48 × 10⁶ m)
  • f₀ = focal length of the objective lens (15 m)
  • R = radius of the lunar orbit (3.8 × 10⁸ m) (Note: the question had “sm” which I assume was a typo for “m”)

Rearranging the equation to solve for Dᵢ:

Dᵢ ≈ (f₀ / R) * Dₘ

Substituting the values:

Dᵢ ≈ (15 m / 3.8 × 10⁸ m) * 3.48 × 10⁶ m

Dᵢ ≈ 0.137 m or 13.7 cm

15. Use the mirror equation to deduce that:

(a) An object placed between f and 2f of a concave mirror produces a real image beyond 2 f.

(b) A convex mirror always produces a virtual image independent of the location of the object.

(c) The virtual image produced by a convex mirror is always diminished in size and is located Between the focus and the pole.

(d) An object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

Ans :

(a)

(b)

(c)

(d)

16. A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab?

Ans : 

The apparent depth (d’) is related to the real depth (d) and the refractive index (n) of the medium by the formula:

d’ = d / n

In this case:

  • d = 15 cm (the thickness of the glass slab, as this is the ‘real depth’ the pin appears to be at)
  • n = 1.5 (refractive index of glass)

So, the apparent depth is:

d’ = 15 cm / 1.5 = 10 cm

3. Calculate how much the pin appears to be raised:

The pin appears to be raised by the difference between the real depth and the apparent depth:

Apparent raise = d – d’ = 15 cm – 10 cm = 5 cm

17. (a) Figure shows a cross-section of’light pipe’ made of a glass fiber of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure.

(b) What is the answer if there is no outer covering of the pipe?

NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Q17

Ans : 

(b)

18. The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?

Ans : 

The maximum possible focal length of the lens is 0.75 m or 75 cm.

19. A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.

Ans : 

20.(a) Determine the ‘effective focal length of the combination of two lenses in question 10, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all?

(b) An object 1.5 cm in size is placed on the side of the convex lens in the above arrangement. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.

Ans : 

(b)

21. At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.

Ans : 

n₁sinθ₁ = n₂sinθ₂

Where:

  • n₁ = refractive index of the prism (1.524)
  • θ₁ = critical angle (θc)
  • n₂ = refractive index of air (approximately 1)
  • θ₂ = 90 degrees

So:

1.524 * sin(θc) = 1 * sin(90°) sin(θc) = 1 / 1.524 θc = arcsin(1/1.524) ≈ 40.8 degrees

2. Relate angles within the prism:

Let:

  • A = refracting angle of the prism (60°)
  • θ₂ = angle of refraction at the first face
  • θc = critical angle (40.8°)

We know that:

θ₂ + θc = A

So:

θ₂ = A – θc = 60° – 40.8° ≈ 19.2°

3. Use Snell’s Law at the first face:

Now we can use Snell’s Law at the first face of the prism to find the angle of incidence (θ₁):

n₀sinθ₁ = n₁sinθ₂

Where: 

  • θ₁ = angle of incidence (what we want to find)

So:

sinθ₁ = 1.524 * sin(19.2°) 

θ₁ = arcsin(1.524 * sin(19.2°)) ≈ 29.8°

22. A card sheet divided into squares each of size 1 mm2 is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 10 cm) held close to the eye.

(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?

(b) What is the angular magnification (magnifying power) of the lens?

(c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.

Ans : 

(b)

(c)

 The linear magnification tells you how much larger or smaller the image is compared to the object, while the angular magnification tells you how much larger the object appears when using the magnifying glass compared to viewing it at the near point with the naked eye.

23. (a) At what distance should the lens be held from the figure in previous question in order to view the squares distinctly with the maximum possible magnifying power?

(b) What is the magnification in this case?

(c) Is the magnification equal to the magnifying power in this case? Explain.

Ans : 

We want to find the object distance (u) for which the image distance (v) is -25 cm (negative because it’s a virtual image on the same side as the object). We’ll use the lens formula:

1/f = 1/v – 1/u

Where:

  • f = focal length = 10 cm
  • v = image distance = -25 cm
  • u = object distance (what we want to find)

1/10 = 1/(-25) – 1/u 1/u = -1/10 – 1/25 1/u = (-5 – 2) / 50 1/u = -7/50 u = -50/7 ≈ -7.14 cm

So, the lens should be held approximately 7.14 cm from the card sheet.

(b) Magnification in this Case:

The magnification (m) is given by:

m = -v/u = -(-25 cm) / (-50/7 cm) = -25 * (-7/50) = 3.5

The magnification is 3.5.

(c) Is Magnification Equal to Magnifying Power?

Yes, in this specific case where the final image is formed at the near point (least distance of distinct vision), the magnitude of the linear magnification is equal to the angular magnification (magnifying power).

24. What should be the distance between the object in previous question and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm2. Would you be able to see the squares distinctly with your eyes very close to the magnifier?

Ans : 

 Find the Required Magnification:

  • The area of the image square is 6.25 mm², and the area of the object square is 1 mm².
  • The area magnification is the ratio of the image area to the object area: 6.25 mm² / 1 mm² = 6.25
  • The linear magnification (m) is the square root of the area magnification: √6.25 = 2.5

25. Answer the following question:

(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?

(b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?

(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?

(d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?

(e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?

Ans : 

(a) Angular Magnification: A magnifying glass lets you bring the object closer to your eye, increasing the angle it subtends. The virtual image subtends the same angle as the close object, which is now a larger angle than if you tried to view the object at that close distance without the lens.

(b) Eye Position: Yes, angular magnification decreases as the eye moves back from the magnifying glass. Maximum magnification is when the eye is as close as possible to the lens.

(c) Magnifying Power Limit: Lens aberrations and manufacturing difficulties limit how small a focal length (and thus how high a magnification) can be effectively used.

(d) Compound Microscope Focal Lengths: Short focal length objective gives high lateral magnification; short focal length eyepiece further magnifies the intermediate image.

(e) Eye Position for Microscope: Eyes should be positioned at the exit pupil (a short distance from the eyepiece) for the full field of view and best image quality.

26. An angular magnification (magnifying power) of 30 X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?

Ans : 

27. A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when:

(a) the telescope is in normal adjustment (i.e., when the final image is at infinity)?

(b) the final image is formed at the least distance of distinct vision (25 cm)?

Ans : 

M = -f₀ / fₑ

Where:

  • f₀ = focal length of the objective lens (140 cm)
  • fₑ = focal length of the eyepiece (5.0 cm)

Substituting the values:

M = -140 cm / 5.0 cm = -28

The magnifying power is 28.

(b) Final Image at Least Distance of Distinct Vision (25 cm):

When the final image is formed at the least distance of distinct vision (D = 25 cm), the magnifying power (M) is given by:

M = -f₀ / fₑ *

M = -140 cm / 5.0 cm * (1 + 5.0 cm / 25 cm) M = -28 * (1 + 0.2) M = -28 * 1.2 M = -33.6

28. (a) For the telescope described in previous question 34 (a), what is the separation between the objective and the eyepiece?

(b) If this telescope is used to view a 100 m tall tower 3 km away. What is the height of the image of the tower formed by the objective lens?

(c) What is the height of the final image of the tower if it is formed at 25 cm?

Ans : 

(b)

(c)

29. A Cassegrain telescope uses two mirrors as shown in figure. Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be?

Ans : 

30.Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in figure. A current in the coil produces a deflection of 3.5° of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?

NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Q37

Ans : 

31. Figure shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid?

NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Q39

Ans :

Now a liquid is filled between lens and plane mirror and the image is formed at position of object at 45 cm. The image is formed on the position of object itself, the object must be placed at focus of equivalent lens of Biconvex of glass and Plano convex lens of liquid

NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments Q39.4
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Dr. Upendra Kant Chaubey
Dr. Upendra Kant Chaubeyhttps://education85.com
Dr. Upendra Kant Chaubey, An exceptionally qualified educator, holds both a Master's and Ph.D. With a rich academic background, he brings extensive knowledge and expertise to the classroom, ensuring a rewarding and impactful learning experience for students.
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