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.
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.
