Online Test — Light – Reflection and Refraction
20 Questions • 15 min • Chapter MCQ
15:00
Question 1 of 20
easy
According to the laws of reflection, the angle of incidence is:
greater than the angle of reflection
equal to the angle of reflection
less than the angle of reflection
twice the angle of reflection
Explanation: The first law of reflection states $\angle i = \angle r$.
Question 2 of 20
easy
The focal length of a spherical mirror is related to its radius of curvature by:
$f = R$
$f = 2R$
$f = \frac{R}{2}$
$f = \frac{R}{4}$
Explanation: For a spherical mirror, $f = \frac{R}{2}$.
Question 3 of 20
easy
Which mirror is used as a reflector in vehicle headlights?
Plane mirror
Convex mirror
Concave mirror
Cylindrical mirror
Explanation: A concave mirror with the bulb at its focus produces a strong parallel beam.
Question 4 of 20
easy
The centre of the reflecting surface of a spherical mirror is called the:
focus
pole
centre of curvature
aperture
Explanation: The geometric centre of the mirror's surface is the pole (P).
Question 5 of 20
easy
A convex mirror always forms an image that is:
real and inverted
virtual, erect and diminished
real and enlarged
virtual and enlarged
Explanation: A convex mirror gives a virtual, erect, diminished image for every object position.
Question 6 of 20
medium
In the New Cartesian sign convention, the focal length of a concave mirror is taken as:
positive
negative
zero
infinite
Explanation: A concave mirror has a negative focal length in the New Cartesian convention.
Question 7 of 20
easy
The mirror formula is:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
$v + u = f$
$\frac{1}{f} = v + u$
Explanation: For mirrors, $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Question 8 of 20
easy
Magnification produced by a mirror is given by:
$m = \frac{v}{u}$
$m = -\frac{v}{u}$
$m = \frac{u}{v}$
$m = v \times u$
Explanation: For mirrors, $m = -\frac{v}{u} = \frac{h'}{h}$.
Question 9 of 20
medium
An object is placed at $20\,\text{cm}$ from a concave mirror of focal length $10\,\text{cm}$ ($u=-20$, $f=-10$). The image distance is:
$+20\,\text{cm}$
$-20\,\text{cm}$
$-40\,\text{cm}$
$-10\,\text{cm}$
Explanation: $\frac{1}{v} = \frac{1}{-10} - \frac{1}{-20} = -\frac{1}{20}$, so $v = -20\,\text{cm}$ (object at C).
Question 10 of 20
medium
A magnification of $+0.5$ means the image is:
real, inverted and diminished
virtual, erect and diminished
virtual, erect and enlarged
real, erect and enlarged
Explanation: Positive $m$ means virtual and erect; $|m| = 0.5 < 1$ means diminished.
Question 11 of 20
medium
A concave mirror forms a real, inverted image of the same size as the object. The object is at the:
focus
pole
centre of curvature
infinity
Explanation: When the object is at C, the image is at C — real, inverted and the same size.
Question 12 of 20
easy
Bending of light as it passes from one transparent medium to another is called:
reflection
dispersion
refraction
scattering
Explanation: The change of direction at the boundary of two media is refraction.
Question 13 of 20
easy
The refractive index of a medium is defined as:
$n = \frac{v}{c}$
$n = \frac{c}{v}$
$n = c \times v$
$n = c - v$
Explanation: Refractive index $n = \frac{c}{v}$.
Question 14 of 20
easy
When light travels from a rarer to a denser medium, it bends:
away from the normal
towards the normal
along the surface
back to the source
Explanation: Light slows down in the denser medium and bends towards the normal.
Question 15 of 20
easy
The lens formula is:
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{u} - \frac{1}{v} = \frac{1}{f}$
$v + u = f$
Explanation: For lenses, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Question 16 of 20
easy
The SI unit of the power of a lens is the:
metre
watt
dioptre
newton
Explanation: Power of a lens, $P = \frac{1}{f}$ (in metres), is measured in dioptre (D).
Question 17 of 20
medium
A convex lens of focal length $50\,\text{cm}$ has a power of:
$+2\,\text{D}$
$+0.5\,\text{D}$
$-2\,\text{D}$
$+50\,\text{D}$
Explanation: $f = 0.5\,\text{m}$, so $P = \frac{1}{0.5} = +2\,\text{D}$.
Question 18 of 20
medium
An object is placed $30\,\text{cm}$ from a convex lens of focal length $15\,\text{cm}$ ($u=-30$, $f=+15$). The image distance is:
$+30\,\text{cm}$
$+15\,\text{cm}$
$-30\,\text{cm}$
$+45\,\text{cm}$
Explanation: $\frac{1}{v} = \frac{1}{15} + \frac{1}{-30} = \frac{1}{30}$, so $v = +30\,\text{cm}$ (object at 2F).
Question 19 of 20
medium
The speed of light in a medium of refractive index $1.5$ is (taking $c = 3 \times 10^8\,\text{m/s}$):
$2 \times 10^8\,\text{m/s}$
$3 \times 10^8\,\text{m/s}$
$4.5 \times 10^8\,\text{m/s}$
$1.5 \times 10^8\,\text{m/s}$
Explanation: $v = \frac{c}{n} = \frac{3 \times 10^8}{1.5} = 2 \times 10^8\,\text{m/s}$.
Question 20 of 20
medium
For a lens, the magnification is given by:
$m = -\frac{v}{u}$
$m = \frac{v}{u}$
$m = \frac{u}{v}$
$m = u \times v$
Explanation: For lenses, $m = \frac{v}{u} = \frac{h'}{h}$ (no minus sign, unlike mirrors).