IMO Practice Test: Light – Reflection and Refraction | Class 10 Physics

Question 1 of 13 hard

A concave mirror of focal length $12\,\text{cm}$ forms a real image twice the size of the object. The object distance is:

$-18\,\text{cm}$

$-24\,\text{cm}$

$-6\,\text{cm}$

$-12\,\text{cm}$

Explanation: Real image, $m=-2 \Rightarrow v=2u$. Then $\frac{1}{v}+\frac{1}{u}=\frac{1}{2u}+\frac{2}{2u}=\frac{3}{2u}=\frac{1}{-12}$, giving $2u=-36$, so $u=-18\,\text{cm}$ (and $v=-36\,\text{cm}$).

Question 2 of 13 medium

Light of speed $3 \times 10^8\,\text{m/s}$ in vacuum travels through a medium at $1.2 \times 10^8\,\text{m/s}$. The refractive index of the medium is:

$1.5$

$2.0$

$2.5$

$0.4$

Explanation: $n = \frac{c}{v} = \frac{3 \times 10^8}{1.2 \times 10^8} = 2.5$.

Question 3 of 13 medium

Two thin lenses of powers $+3\,\text{D}$ and $-1\,\text{D}$ are kept in contact. The power of the combination is:

$+4\,\text{D}$

$+2\,\text{D}$

$-3\,\text{D}$

$+3\,\text{D}$

Explanation: Powers in contact add: $P = 3 + (-1) = +2\,\text{D}$.

Question 4 of 13 medium

An object is at the focus of a concave mirror. The image is formed at:

the focus

the centre of curvature

infinity

the pole

Explanation: Rays from an object at F become parallel after reflection, so the image is at infinity.

Question 5 of 13 medium

A convex lens forms a virtual, erect and enlarged image. The object must be placed:

beyond 2F

between F and the optical centre

at 2F

at F

Explanation: Only when the object is between F and O does a convex lens form a virtual, erect, magnified image (magnifying-glass action).

Question 6 of 13 hard

A concave mirror has $f = -10\,\text{cm}$. An object placed at $u=-10\,\text{cm}$ produces an image:

at $-10\,\text{cm}$

at $-20\,\text{cm}$

at infinity

at $+10\,\text{cm}$

Explanation: Object at F: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-10} - \frac{1}{-10} = 0$, so $v \to \infty$.

Question 7 of 13 medium

A lens has a power of $-2.5\,\text{D}$. Its focal length and type are:

$+40\,\text{cm}$, convex

$-40\,\text{cm}$, concave

$-25\,\text{cm}$, concave

$+25\,\text{cm}$, convex

Explanation: $f = \frac{1}{P} = \frac{1}{-2.5} = -0.4\,\text{m} = -40\,\text{cm}$; negative power means a concave lens.

Question 8 of 13 medium

A ray of light travels from glass ($n=1.5$) into air. Compared with glass, in air the light:

slows down and bends towards the normal

speeds up and bends away from the normal

speeds up and bends towards the normal

keeps the same speed

Explanation: Air is rarer, so light speeds up and bends away from the normal.

Question 9 of 13 hard

An object 2 cm tall is placed 10 cm from a convex lens of focal length 20 cm. The image is:

real, inverted, 4 cm

virtual, erect, 4 cm

real, erect, 2 cm

virtual, inverted, 1 cm

Explanation: $\frac{1}{v} = \frac{1}{20} + \frac{1}{-10} = -\frac{1}{20}$, so $v=-20$; $m = \frac{v}{u} = \frac{-20}{-10} = +2$, image $= 2 \times 2 = 4\,\text{cm}$, virtual and erect.

Question 10 of 13 hard

A plane mirror is rotated by $10^\circ$ while the incident ray is fixed. The reflected ray turns by:

$5^\circ$

$10^\circ$

$20^\circ$

$0^\circ$

Explanation: When a plane mirror turns by an angle $\theta$ with a fixed incident ray, the reflected ray turns by $2\theta = 20^\circ$.

Question 11 of 13 hard

A convex mirror of focal length $15\,\text{cm}$ has an object at $u=-30\,\text{cm}$. The magnification is:

$+\frac{1}{3}$

$-\frac{1}{3}$

$+\frac{1}{2}$

$-2$

Explanation: $\frac{1}{v} = \frac{1}{15} - \frac{1}{-30} = \frac{3}{30} = \frac{1}{10}$, so $v=+10$; $m = -\frac{v}{u} = -\frac{10}{-30} = +\frac{1}{3}$.

Question 12 of 13 medium

For the same object distance, which combination gives a real, magnified image?

convex mirror

concave lens

concave mirror with object between F and C

plane mirror

Explanation: A concave mirror with the object between F and C forms a real, inverted, magnified image; convex mirrors and concave lenses give only virtual diminished images.

Question 13 of 13 hard

A convex lens of focal length 10 cm and a concave lens of focal length 10 cm are placed in contact. The combination behaves as:

a convex lens of f = 5 cm

a concave lens of f = 5 cm

a plane glass plate (no converging or diverging)

a convex lens of f = 20 cm

Explanation: $P = \frac{1}{0.1} + \frac{1}{-0.1} = 0$, so the combination has zero power and acts like a plane plate.

IMO Practice Test — Light – Reflection and Refraction