Reflection of Light: Mirrors — Hard Practice Quiz
A Physics cheat sheet for Reflection of Light: Mirrors — every key formula with its symbols defined — plus a hard-level practice quiz to test recall.
Formulas & key concepts
Law of Reflection: The angle of incidence \(\theta_i\) equals the angle of reflection \(\theta_r\).
Focal Length (Concave Mirror): Focal length \(f\) is half the radius of curvature \(R\).
Focal Length (Convex Mirror): Focal length \(f\) is negative half the radius of curvature \(R\).
Mirror Equation: Relates object distance \(d_o\), image distance \(d_i\), and focal length \(f\).
Magnification Equation: Relates magnification \(m\), image height \(h_i\), object height \(h_o\), and distances.
Practice quiz
An object is placed in front of a concave mirror. A real, inverted image is formed, whose height is one-third the object's height. If the mirror's radius of curvature is $R$, what is the object distance $d_o$ in terms of $R$?
- $\frac{R}{2}$
- $R$
- $2R$
- $\frac{3R}{2}$
Answer: $2R$
A convex mirror forms a virtual image that is one-fourth the height of the object. If the object is placed at a distance $D$ from the mirror, what is the magnitude of the focal length $f$ of the mirror in terms of $D$?
- $\frac{D}{2}$
- $\frac{D}{3}$
- $\frac{2D}{3}$
- $D$
Answer: $\frac{D}{3}$
An object is placed at a distance $d_o$ from a concave mirror with focal length $f$. If the image formed is real, inverted, and twice the size of the object, what is the object distance $d_o$ in terms of $f$?
- $\frac{f}{2}$
- $f$
- $\frac{3f}{2}$
- $2f$
Answer: $\frac{3f}{2}$
A concave mirror has a radius of curvature $R$. An object is placed at a distance $R/4$ from the mirror. What is the magnification $m$ of the image formed?
- $-2$
- $-\frac{1}{2}$
- $\frac{1}{2}$
- $2$
Answer: $2$
A light ray strikes a plane mirror at an angle of incidence $\theta_i$. If the mirror is replaced by a concave mirror with focal length $f$, and an object is placed at a distance $d_o = 3f$, how does the image height $h_i$ relate to the object height $h_o$?
- $h_i = -h_o$
- $h_i = -2h_o$
- $h_i = -\frac{h_o}{2}$
- $h_i = \frac{h_o}{2}$
Answer: $h_i = -\frac{h_o}{2}$
An object is placed in front of a concave mirror. The image formed is virtual and three times the size of the object. If the object is moved to a new position such that the image formed is real and half the size of the object, what is the ratio of the initial object distance $d_{o1}$ to the final object distance $d_{o2}$?
- $\frac{1}{3}$
- $\frac{2}{3}$
- $\frac{2}{9}$
- $\frac{3}{2}$
Answer: $\frac{2}{9}$
A convex mirror has a radius of curvature $R$. An object is placed at a distance $d_o$ from the mirror. If the image formed is virtual and has a magnification $m$, derive an expression for $d_o$ in terms of $R$ and $m$.
- $\frac{R(m-1)}{2m}$
- $\frac{R(1-m)}{2m}$
- $\frac{R(m+1)}{2m}$
- $\frac{R(1+m)}{2m}$
Answer: $\frac{R(1-m)}{2m}$
An object is placed at a distance $d_o$ from a concave mirror. The image formed is real and has a height $h_i$. If the object height is $h_o$, and the mirror's focal length is $f$, derive an expression for $d_o$ in terms of $f$, $h_i$, and $h_o$. (Assume $h_i$ is the signed height, so for a real image, $h_i$ is negative.)
- $f \frac{h_o - h_i}{h_i}$
- $f \frac{h_i - h_o}{h_i}$
- $f \frac{h_o}{h_i - h_o}$
- $f \frac{h_i}{h_o - h_i}$
Answer: $f \frac{h_i - h_o}{h_i}$
A concave mirror has a focal length $f$. An object is placed at a distance $d_o$ from the mirror. If the image formed is virtual and its height is $N$ times the object's height, what is the object distance $d_o$ in terms of $f$ and $N$?
- $f \frac{N+1}{N}$
- $f \frac{N-1}{N}$
- $f \frac{N}{N-1}$
- $f \frac{N}{N+1}$
Answer: $f \frac{N-1}{N}$
Consider two mirrors, a concave mirror (M1) and a convex mirror (M2), both with the same absolute radius of curvature $R$. An object is placed at a distance $R/2$ from M1 and at a distance $R$ from M2. Compare the nature and magnification of the images formed.
- M1: Real, inverted, $m=-1$; M2: Virtual, upright, $m=1/3$.
- M1: Virtual, upright, $m=2$; M2: Real, inverted, $m=-1/3$.
- M1: Real, inverted, infinitely magnified; M2: Virtual, upright, $m=1/3$.
- M1: Virtual, upright, infinitely magnified; M2: Real, inverted, $m=-1/3$.
Answer: M1: Real, inverted, infinitely magnified; M2: Virtual, upright, $m=1/3$.
Select a subject
Select a subject from the left panel to begin exploring formulas.