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JEE Advanced Ray Optics and Optical Instruments — practice questions

Explore 88 free Ray Optics and Optical Instruments questions in JEE Advanced 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 88.

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 →The critical angle for total internal reflection at a glass-air interface is $42^\circ$. The refractive index of the glass is approximately:Read the answer and worked solution →An object is placed $30\,\text{cm}$ in front of a concave mirror of focal length $20\,\text{cm}$. The image distance from the mirror is:Read the answer and worked solution →Two plane mirrors are inclined at an angle of $60^\circ$. The number of images of a point object placed between them is:Read the answer and worked solution →A convex mirror always forms an image of a real object that is:Read the answer and worked solution →A point source of light is placed at the origin in an isotropic medium. The wavefronts associated with the light are:Read the answer and worked solution →The mirror equation relates u, v and f asRead the answer and worked solution →The power P of a lens of focal length f (in metres) is given byRead the answer and worked solution →An object is placed 30 cm from a concave mirror of focal length 20 cm. The image distance isRead the answer and worked solution →Refractive index of water with respect to air is 4/3. The critical angle for a water–air interface is approximatelyRead the answer and worked solution →A convex lens of focal length 20 cm is placed in contact with a concave lens of focal length 25 cm. The power of the combination (in dioptres) isRead the answer and worked solution →Light passes from air (n=1) into glass (n=1.5) at an incidence angle of 60°. Using Snell's law, the angle of refraction isRead the answer and worked solution →For a thin prism of small refracting angle A and refractive index n, the deviation δ is approximatelyRead the answer and worked solution →An astronomical telescope in normal adjustment has an objective of focal length 100 cm and an eyepiece of focal length 5 cm. Its angular magnification isRead the answer and worked solution →The apparent depth of an object lying at the bottom of a pool of water (n = 4/3) is, in terms of its real depth h,Read the answer and worked solution →A thin prism of refracting angle 5° is made of glass with n = 1.5. The deviation δ of a light ray passing through it is approximatelyRead the answer and worked solution →Two thin convex lenses, each of focal length 20 cm, are placed coaxially 10 cm apart in air. The equivalent focal length of the combination isRead the answer and worked solution →Refraction — true or false?Read the answer and worked solution →Find the property that does not match concave mirrorsRead the answer and worked solution →Complete the lens formulaRead the answer and worked solution →Cartesian sign convention: positive or negative?Read the answer and worked solution →True or false: the sign conventionRead the answer and worked solution →Complete the condition for total internal reflectionRead the answer and worked solution →Tap every example of total internal reflectionRead the answer and worked solution →Critical angleRead the answer and worked solution →True or false: optical fibresRead the answer and worked solution →Match each formula to what it describesRead the answer and worked solution →Converging or diverging?Read the answer and worked solution →Complete the definition of lens powerRead the answer and worked solution →Lenses in contactRead the answer and worked solution →

Worked practice questions

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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The critical angle for total internal reflection at a glass-air interface is $42^\circ$. The refractive index of the glass is approximately:

  1. $1.33$
  2. $1.41$
  3. $1.49$
  4. $2.42$
Show answer and solution

Correct answer: C. $1.49$

For total internal reflection at a denser-to-rarer interface, $\mu = \dfrac{1}{\sin C}$. With $C = 42^\circ$, $\sin 42^\circ \approx 0.669$, so $\mu = 1/0.669 \approx 1.49$, which matches typical crown glass. **Option A is water and option D is diamond.**

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An object is placed $30\,\text{cm}$ in front of a concave mirror of focal length $20\,\text{cm}$. The image distance from the mirror is:

  1. $+12\,\text{cm}$
  2. $-60\,\text{cm}$
  3. $+60\,\text{cm}$
  4. $-12\,\text{cm}$
Show answer and solution

Correct answer: B. $-60\,\text{cm}$

Using the standard sign convention, $u = -30\,\text{cm}$ and $f = -20\,\text{cm}$ for a concave mirror. The mirror formula $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$ gives $\dfrac{1}{v} = \dfrac{1}{-20} - \dfrac{1}{-30} = -\dfrac{1}{60}$, so $v = -60\,\text{cm}$. The negative sign indicates the image is real and on the same side as the object, $60\,\text{cm}$ in front of the mirror.

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