A small change in an object's position, a mirror's movement or the surrounding medium can alter where an image forms. Ray Optics helps you understand these changes using ray diagrams, geometrical relationships and mathematical formulas.
Physics Wallah offers Ray Optics and Optical Instruments Important Questions to support your preparation. Use these questions to review important concepts, understand formula applications and practise solving numerical problems while avoiding common sign-convention mistakes.
Answer: The angle of incidence equals the angle of reflection, and the incident ray, reflected ray and normal lie in the same plane.
For an angle of incidence (i):
Angle between the incident and reflected rays = 2i
Deviation = 180° − 2i
If the incident ray makes an angle α with the mirror surface, then:
Deviation = 2α
At normal incidence, deviation is 180° because the ray retraces its path.
Answer:
If the mirror rotates through an angle θ while the incident ray remains fixed, the reflected ray rotates through 2θ.
If the object moves a distance x while the mirror remains fixed, its image moves by x in the opposite direction.
If the mirror moves a distance x towards or away from a stationary object along the normal, the image moves by 2x in the same direction as the mirror.
Answer: The image of a real object is virtual, erect, the same size as the object and laterally inverted. It forms as far behind the mirror as the object is in front.
The minimum mirror height required to see the complete image of a person of height h is:
Minimum mirror height = h/2
Removing part of the mirror does not change the image’s size or position, although less of the reflected light may reach the observer.
Answer: For two plane mirrors inclined at an angle θ, begin by calculating:
N = 360°/θ
If N is an integer:
If N is even, the number of images is N − 1.
If N is odd, the number is generally N − 1 when the object lies on the angle bisector; otherwise, it is N.
If N is not an integer, the number of images is generally the greatest integer less than N.
The exact result depends on the object’s position and the mirror arrangement.
Answer: Using the Cartesian sign convention for spherical mirrors:
1/v + 1/u = 1/f
Magnification: m = −v/u = hᵢ/hₒ
Here, u is object distance, v is image distance and f is focal length.
For a spherical mirror:
f = R/2
A positive magnification indicates an erect image, while a negative magnification indicates an inverted image.
Answer:
Concave mirror: An object between the pole and focus produces a virtual, erect and enlarged image. An object beyond the focus generally produces a real, inverted image.
Convex mirror: A real object always produces a virtual, erect and diminished image.
For a concave mirror, an object at the centre of curvature forms an image of the same size at the centre of curvature.
Answer: Refraction is the change in the direction of light as it passes between media because its speed changes.
The absolute refractive index is:
μ = c/v
Snell’s law is:
μ₁ sin i = μ₂ sin r
Frequency remains unchanged during refraction, while speed and wavelength change.
When light enters an optically denser medium, its speed and wavelength decrease, and it bends towards the normal.
Answer: For an object in a medium viewed nearly normally from another medium:
Apparent depth = Real depth × (μ observer/μ object)
For an object in water viewed from air:
Apparent depth ≈ Real depth/μ water
The apparent shift is:
Apparent shift = Real depth − Apparent depth
For normal dispersion in a transparent medium, violet light generally has a higher refractive index than red light. Consequently, violet light has a smaller apparent depth and a greater apparent shift.
Answer: Total internal reflection occurs only when:
Light travels from an optically denser medium to an optically rarer medium.
The angle of incidence exceeds the critical angle.
At the critical angle, the refracted ray travels along the boundary.
The critical-angle formula is:
sin C = μ rarer/μ denser
For an optical fibre, the refractive index of the core must be greater than that of the cladding. Total internal reflection keeps light travelling through the core.
Answer: For a thin lens in a surrounding medium:
1/f = (μ lens/μ medium − 1)(1/R₁ − 1/R₂)
The focal length depends on the refractive index of the lens relative to the surrounding medium and on the radii of curvature.
A convex lens is converging when its refractive index is greater than that of the surrounding medium. If both indices are equal, the lens has no effective optical power.
Answer: Using the Cartesian sign convention for thin lenses:
1/v − 1/u = 1/f
Magnification: m = v/u = hᵢ/hₒ
A convex lens can form real, inverted images or a virtual, erect, enlarged image when the object lies between the optical centre and the focus.
A concave lens forms a virtual, erect and diminished image for a real object.
For thin lenses in contact:
P net = P₁ + P₂ + P₃ + …
For two thin lenses separated by distance d in air:
P net = P₁ + P₂ − dP₁P₂
Here, powers are measured in dioptres and d in metres.
Answer: The effect depends on the orientation of the cut.
A cut along the principal axis generally leaves the focal length and power of each resulting piece unchanged, although the aperture and brightness decrease.
A cut perpendicular to the principal axis changes the curvature arrangement of the pieces. For a symmetric biconvex lens cut through its optical centre, each half has twice the original focal length and half the original power.
Do not apply the second result to every lens shape without checking its geometry.
Answer: For a prism of angle A:
Deviation, δ = i + e − A
At minimum deviation, δₘ:
Angle of incidence equals angle of emergence: i = e
Refraction angles inside the prism are equal: r₁ = r₂ = A/2
The refractive index is:
μ = sin[(A + δₘ)/2]/sin(A/2)
For a thin prism, the approximate deviation in radians is:
δ ≈ A(μ − 1)
Answer: A simple microscope uses a convex lens to produce a magnified virtual image.
For a final image at infinity:
M = D/f
For a final image at the near point:
M = 1 + D/f
Here, D is the least distance of distinct vision, usually taken as 25 cm, and f is the focal length of the lens.
Answer: A compound microscope uses an objective lens and an eyepiece.
Total magnifying power = Objective magnification × Eyepiece magnification
For normal adjustment, the approximate magnitude is:
M = (L/fₒ)(D/fₑ)
where L is the tube length, fₒ is the objective focal length, fₑ is the eyepiece focal length and D is the least distance of distinct vision.
Answer: For an astronomical telescope in normal adjustment:
Magnifying power (magnitude) = fₒ/fₑ
Telescope length = fₒ + fₑ
Here, fₒ and fₑ are the focal lengths of the objective and eyepiece.
A Galilean telescope uses a convex objective and a concave eyepiece. Its length in normal adjustment is:
Length = fₒ − |fₑ|
Plane-mirror deviation: 180° − 2i
Spherical mirror: 1/v + 1/u = 1/f
Mirror magnification: m = −v/u
Refractive index: μ = c/v
Snell’s law: μ₁ sin i = μ₂ sin r
Apparent depth from air: Real depth/μ
Critical angle: sin C = μ rarer/μ denser
Thin lens: 1/v − 1/u = 1/f
Lens power in contact: P net = ΣP
Prism deviation: δ = i + e − A
Simple microscope: M = D/f (final image at infinity)
Astronomical telescope: M = fₒ/fₑ (magnitude, normal adjustment)
For effective NEET revision, practise ray diagrams and numerical questions alongside formula recall. Pay particular attention to sign conventions, image movement, critical-angle conditions and optical-instrument magnification. Physics Wallah’s Physics Important Questions can help you reinforce these concepts and prepare for application-based questions.
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