Class 10 Light — Reflection and Refraction Mock Test (CBSE)

Topics in Light — Reflection and Refraction

• Reflection of light
• Spherical mirrors
• Image formation by spherical mirrors and ray diagrams
• Sign convention for spherical mirrors
• Mirror formula and magnification
• Refraction of light
• Refraction through a rectangular glass slab
• Refractive index
• Refraction by spherical lenses
• Image formation by lenses and ray diagrams
• Sign convention for spherical lenses
• Lens formula and magnification
• Power of a lens

Light — Reflection and Refraction practice questions with answers

1. Q1Multiple choice·1 mark
   A concave lens has a focal length of $-50\text{ cm}$. What is the power of this lens?

   1. (a)$+2\text{ D}$
   2. (b)$-2\text{ D}$
   3. (c)$+0.5\text{ D}$
   4. (d)$-0.5\text{ D}$
   Show answer and step marks

   1. Correct option: (b) $-2\text{ D}$. Using the power formula $P = \frac{1}{f\text{ (in m)}} = \frac{1}{-0.5\text{ m}} = -2\text{ D}$.+1

   Answer: (b)
2. Q2Multiple choice·1 mark
   If the refractive index of a medium with respect to vacuum is $1.5$, what is the speed of light in this medium? (Take speed of light in vacuum as $3 \times 10^8\text{ m/s}$)

   1. (a)$2 \times 10^8\text{ m/s}$
   2. (b)$4.5 \times 10^8\text{ m/s}$
   3. (c)$1.5 \times 10^8\text{ m/s}$
   4. (d)$3 \times 10^8\text{ m/s}$
   Show answer and step marks

   1. Correct option: (a) $2 \times 10^8\text{ m/s}$. Using the relation $v = \frac{c}{n} = \frac{3 \times 10^8}{1.5} = 2 \times 10^8\text{ m/s}$.+1

   Answer: (a)
3. Q3Multiple choice·1 mark
   The absolute refractive indices of medium A and medium B are $1.5$ and $2.0$ respectively. The refractive index of medium B with respect to medium A is:

   1. (a)$0.75$
   2. (b)$1.33$
   3. (c)$3.0$
   4. (d)$1.5$
   Show answer and step marks

   1. Correct option: (b) $1.33$+1

   Answer: (b)
4. Q4Multiple choice·1 mark
   An object is placed at a distance of $15\text{ cm}$ in front of a concave mirror of focal length $10\text{ cm}$. The nature and position of the image formed are:

   1. (a)Real, inverted, at $30\text{ cm}$ in front of the mirror
   2. (b)Virtual, erect, at $30\text{ cm}$ behind the mirror
   3. (c)Real, inverted, at $6\text{ cm}$ in front of the mirror
   4. (d)Virtual, erect, at $6\text{ cm}$ behind the mirror
   Show answer and step marks

   1. Correct option: (a) Real, inverted, at $30\text{ cm}$ in front of the mirror+1

   Answer: (a)
5. Q5Short answer·2 marks
   A doctor prescribes a corrective lens of power $+2.5\text{ D}$. (a) What type of lens is this and which defect of vision does it correct? (b) Calculate the focal length of this lens in centimetres.
   Show answer and step marks

   1. (a) Type: convex (converging) lens, since the power is positive.+0.5
   2. (a) Defect corrected: hypermetropia (long-sightedness); presbyopia is also accepted.+0.5
   3. (b) Calculation of focal length: $f = \frac{1}{P} = \frac{1}{+2.5\text{ D}} = +0.4\text{ m} = +40\text{ cm}$.+1

   Answer: (a) Convex lens; it corrects hypermetropia (presbyopia also accepted). (b) Focal length $= +40\text{ cm}$
6. Q6Short answer·3 marks
   An object of height $4.0\text{ cm}$ is placed at a distance of $25.0\text{ cm}$ in front of a concave mirror of focal length $15.0\text{ cm}$. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Find the nature and the height of the image.
   Show answer and step marks

   1. State the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ and substitute given values with sign conventions ($u = -25.0\text{ cm}$, $f = -15.0\text{ cm}$).+1
   2. Calculate image distance: $\frac{1}{v} = \frac{1}{-15.0} - \frac{1}{-25.0} = -\frac{2}{75}$, giving $v = -37.5\text{ cm}$. The screen should be placed $37.5\text{ cm}$ in front of the mirror.+1
   3. Calculate magnification $m = -\frac{v}{u} = -\frac{-37.5}{-25.0} = -1.5$ and height $h' = m \times h = -1.5 \times 4.0 = -6.0\text{ cm}$, so the image is $6.0\text{ cm}$ tall.+0.5
   4. Nature: the image is real and inverted (and magnified), as $v$ and $m$ are negative.+0.5

   Answer: Screen at $37.5\text{ cm}$ in front of the mirror; the image is real, inverted and magnified, of height $6.0\text{ cm}$ ($h' = -6.0\text{ cm}$).

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