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KCET Ray Optics and Optical Instruments — practice questions

Explore 56 free Ray Optics and Optical Instruments questions in KCET Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Ray Optics and Optical Instruments questions and exercises with answers

Showing questions 1–30 of 56.

A convex lens forms a real image at $60$ cm on one side of an object placed $20$ cm in front of it. Find the focal length of the lens.Read the answer and worked solution →Two thin lenses of focal lengths $+10$ cm and $+20$ cm are placed in contact. The power of the combination, in diopters, is:Read the answer and worked solution →A light ray travels from glass ($\mu = 1.5$) to air. The critical angle for total internal reflection at the glass–air interface is approximately:Read the answer and worked solution →Snell's law of refraction states:Read the answer and worked solution →For a plane mirror, image is:Read the answer and worked solution →Focal length of a concave mirror with radius of curvature R:Read the answer and worked solution →Mirror formula relating object distance u, image distance v, focal length f:Read the answer and worked solution →Magnification by a mirror:Read the answer and worked solution →Refractive index of water is 1.33. Critical angle for water-air interface:Read the answer and worked solution →For a thin lens, lens-maker's formula:Read the answer and worked solution →For a convex lens, an object at 2f forms image at:Read the answer and worked solution →Power of a lens (P) is defined as:Read the answer and worked solution →Two thin lenses of powers P₁ and P₂ in contact give combined power:Read the answer and worked solution →Refraction by a prism: deviation δ depends on:Read the answer and worked solution →In a compound microscope, magnification by objective is m_o, eyepiece is m_e. Total magnification:Read the answer and worked solution →Telescope's angular magnification (normal adjustment, image at infinity):Read the answer and worked solution →Speed of light in a medium of refractive index n:Read the answer and worked solution →Object at focus of a convex lens. Image is:Read the answer and worked solution →An object 4 cm tall is placed 30 cm in front of a concave mirror of focal length 20 cm. Image position and size:Read the answer and worked solution →For an apparent depth of swimming pool seen from above (real depth d, water n = 4/3):Read the answer and worked solution →Critical angle for diamond (n = 2.42):Read the answer and worked solution →Combination of two lenses, f₁ = 20 cm convex and f₂ = -25 cm concave, in contact. Equivalent power:Read the answer and worked solution →Dispersion of light by prism: angular dispersion (δ_violet - δ_red) for thin prism:Read the answer and worked solution →A coin at the bottom of a 0.5 m water tank (n = 4/3) appears at depth:Read the answer and worked solution →In Young's double slit experiment, fringe width:Read the answer and worked solution →Refractive index of medium 1 w.r.t. medium 2 is n₁₂. Then:Read the answer and worked solution →Magnifying power of simple magnifier (image at near point):Read the answer and worked solution →For object at 60 cm from convex lens of f = 20 cm, image is:Read the answer and worked solution →For total internal reflection at glass-water interface (n_g = 1.5, n_w = 1.33), critical angle:Read the answer and worked solution →A ray enters a glass slab at 45°. Glass n = 1.5. Angle of refraction inside glass:Read the answer and worked solution →

Worked practice questions

A convex lens forms a real image at $60$ cm on one side of an object placed $20$ cm in front of it. Find the focal length of the lens.

  1. $10$ cm
  2. $12$ cm
  3. $15$ cm
  4. $20$ cm
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Correct answer: C. $15$ cm

Lens formula: $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$. Sign convention: $u = -20$ cm (object on the left), $v = +60$ cm (real image on the right). We have $\dfrac{1}{f} = \dfrac{1}{60} - \dfrac{1}{-20}$. That is $\dfrac{1}{f} = \dfrac{1}{60} + \dfrac{3}{60}$. That is $\dfrac{1}{f} = \dfrac{4}{60}$. So $\dfrac{1}{f} = \dfrac{1}{15}$. So $f = 15$ cm. **Trap.** Forgetting the sign on $u$ gives $\dfrac{1}{f} = \dfrac{1}{60} - \dfrac{1}{20} = -\dfrac{1}{30}$, leading to a negative $f$ — that's how option B (12) or A (10) creeps in.

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Two thin lenses of focal lengths $+10$ cm and $+20$ cm are placed in contact. The power of the combination, in diopters, is:

  1. $5$ D
  2. $10$ D
  3. $15$ D
  4. $20$ D
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Correct answer: C. $15$ D

**For thin lenses in contact**, powers add: $P = P_1 + P_2$. $P_1 = \dfrac{1}{f_1\text{ (m)}} = \dfrac{1}{0.10} = 10$ D. $P_2 = \dfrac{1}{0.20} = 5$ D. $P_{\text{total}} = 10 + 5 = 15$ D. Equivalently, $\dfrac{1}{F} = \dfrac{1}{10} + \dfrac{1}{20} = \dfrac{3}{20}$ giving $F = \dfrac{20}{3}$ cm $\approx 6.67$ cm so $P = \dfrac{100}{6.67} = 15$ D.

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A light ray travels from glass ($\mu = 1.5$) to air. The critical angle for total internal reflection at the glass–air interface is approximately:

  1. $30°$
  2. $42°$
  3. $60°$
  4. $90°$
Show answer and solution

Correct answer: B. $42°$

**Critical angle formula** (denser → rarer): $\sin C = \dfrac{\mu_{\text{rarer}}}{\mu_{\text{denser}}} = \dfrac{1}{\mu_{\text{denser}}} = \dfrac{1}{1.5} \approx 0.667$. $C = \sin^{-1}(0.667) \approx 41.8° \approx 42°$. **Why option C (60°) is tempting.** Some recall $\sin^{-1}(0.866) = 60°$ — but that's $\mu = 1/0.866 \approx 1.15$, not $1.5$. **A useful sanity check:** as $\mu$ rises, the critical angle drops (denser glass holds light in more aggressively). Diamond's $C \approx 24°$ — that's why it sparkles.

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