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Ray Optics and Optical Instruments — NEET & JEE MCQs with answers
Ray optics treats light as travelling in straight lines (rays), valid because the wavelength of light is very small compared with the size of ordinary objects. The chapter covers reflection at spherical mirrors, refraction at plane and spherical surfaces and by lenses, total internal reflection, refraction through a prism, and optical instruments (microscopes and telescopes). Key formulae include the mirror equation, lens formula, and lens maker's formula.
Topics in this chapter
- 9.1 Introduction
- 9.2 Reflection of Light by Spherical Mirrors
- 9.3 Refraction
- 9.4 Total Internal Reflection
- 9.5 Refraction at Spherical Surfaces and by Lenses
- 9.6 Refraction Through a Prism
- 9.7 Optical Instruments
Free practice questions
A lens of power $+10\ \text{D}$ has a focal length of:
- A. $+100\ \text{cm}$
- B. $+10\ \text{cm}$
- C. $+1\ \text{cm}$
- D. $+0.1\ \text{cm}$
Show answer
Answer: B — $f=\dfrac{1}{10}=0.10\ \text{m}=+10\ \text{cm}$.
NCERT Class 12 Physics Ch 9 ("The power of a lens of focal length of 1 metre is one dioptre")
A simple magnifier of focal length $12.5$ cm forms the image at the near point ($D = 25$ cm). Its magnification is
- A. $2$
- B. $4$
- C. $2.5$
- D. $3$
Show answer
Answer: D — $m = 1 + \dfrac{D}{f} = 1 + \dfrac{25}{12.5} = 1 + 2 = 3$.
NCERT Class 12 Physics Ch 9 ("the closest comfortable distance for viewing the image is when it is at the near point")
A concave mirror of focal length 7.5 cm forms a magnified real inverted image with |m| = 3 of a real object. The object distance is:
- A. $-22.5\ \text{cm}$
- B. $-30\ \text{cm}$
- C. $-10\ \text{cm}$
- D. $-5\ \text{cm}$
Show answer
Answer: C — For a real inverted image m = −3 = −v/u ⇒ v = 3u. Mirror eq: 1/(3u) + 1/u = 1/f = 1/(−7.5) ⇒ 4/(3u) = −1/7.5 ⇒ u = −10 cm (image at v = −30 cm).
NCERT Class 12 Physics Ch 9 ("f = -15/2 cm = -7.5 cm")
A ray in medium 1 (n₁ = 1) strikes the interface with medium 2 (n₂ = 1.5) at an angle of incidence 60°. The angle of refraction is:
- A. $90^\circ$
- B. $42.26^\circ$
- C. $35.26^\circ$
- D. $60^\circ$
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Answer: C — Snell's law n₁ sin i = n₂ sin r ⇒ sin r = (1 × sin 60°)/1.5 = 0.58, so r = 35.26°. Since n₂ > n₁ the ray bends towards the normal (r < i).
NCERT Class 12 Physics Ch 9 ("Snell's law of refraction")
At minimum deviation, a prism has angle of incidence $60^\circ$ and refracting angle $60^\circ$. The minimum deviation is:
- A. $60^\circ$
- B. $50^\circ$
- C. $45^\circ$
- D. $30^\circ$
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Answer: A — $D_m=2i-A=2(60^\circ)-60^\circ=60^\circ$.
NCERT Class 12 Physics Ch 9 ("where Dm is the angle of minimum deviation")
Which comparison best explains why light can be modelled as straight-line rays for everyday objects (size ~1 cm)?
- A. The wavelength is about ten times larger than the object
- B. The wavelength (~500 nm) is about ten thousand times smaller than the object
- C. The wavelength is exactly one centimetre
- D. The wavelength is about the same as the object size
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Answer: B — A visible wavelength ~500 nm = 5×10⁻⁷ m is ~10⁴ times smaller than a 1 cm (10⁻² m) object, so the wave effectively travels along a straight ray.
NCERT Class 12 Physics Ch 9 ("very small compared to the size of ordinary objects")
At exactly the critical angle of incidence, the refracted ray:
- A. Vanishes completely
- B. Bends towards the normal
- C. Is reflected back at 0°
- D. Travels along (grazes) the interface, i.e. the angle of refraction is 90°
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Answer: D — NCERT: the critical angle corresponds to an angle of refraction of 90°, where the refracted ray grazes the surface.
NCERT Class 12 Physics Ch 9 ("is called the critical angle")
A simple microscope has focal length $f = 2.5$ cm. With the image at the near point ($D = 25$ cm), its magnification is
- A. $11$
- B. $9$
- C. $10$
- D. $12.5$
Show answer
Answer: A — Near-point magnification $m = 1 + \dfrac{D}{f} = 1 + \dfrac{25}{2.5} = 1 + 10 = 11$.
NCERT Class 12 Physics Ch 9 ("the closest comfortable distance for viewing the image is when it is at the near point")
Refraction of light is the change in its direction of propagation when it:
- A. Reflects off a mirror
- B. Is absorbed by an opaque body
- C. Travels in a straight line in one medium
- D. Passes obliquely from one transparent medium into another
Show answer
Answer: D — NCERT: when an obliquely incident ray enters another transparent medium, its direction changes at the interface — this is refraction.
NCERT Class 12 Physics Ch 9 ("The incident ray, the refracted ray and the")
A three-lens combination produces individual magnifications $m_1=+3$, $m_2=+2$ and $m_3=-1$. The total magnification is:
- A. $+5$
- B. $+6$
- C. $-1$
- D. $-6$
Show answer
Answer: D — $m=m_1\,m_2\,m_3=(+3)(+2)(-1)=-6$.
NCERT Class 12 Physics Ch 9 ("the total magnification m of the combination is a product of magnification")
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