101. In mirror problems, the magnification relation \(m=-\frac{v}{u}\) is useful because it gives
ⓐ. relative image size and orientation
ⓑ. only the focal length of the mirror
ⓒ. only the radius of curvature
ⓓ. only the angle of reflection
Correct Answer: relative image size and orientation
Explanation: Mirror magnification compares image height with object height. The relation is \(m=\frac{h_i}{h_o}=-\frac{v}{u}\). The magnitude of \(m\) tells whether the image is enlarged, diminished, or the same size. The sign of \(m\) tells orientation: positive magnification means erect image, while negative magnification means inverted image. Magnification is therefore not only a size number; it also carries orientation information through its sign.
102. A concave mirror has focal length \(-15\,cm\). An object is placed \(30\,cm\) in front of it. The image position is
ⓐ. \(+30\,cm\)
ⓑ. \(-30\,cm\)
ⓒ. \(-10\,cm\)
ⓓ. \(+10\,cm\)
Correct Answer: \(-30\,cm\)
Explanation: \( \textbf{Given data:} \)
Concave mirror focal length is \(f=-15\,cm\).
Object distance is \(u=-30\,cm\).
Use the mirror formula:
\[
\frac{1}{f}=\frac{1}{v}+\frac{1}{u}
\]
Substituting,
\[
\frac{1}{-15}=\frac{1}{v}+\frac{1}{-30}
\]
\[
\frac{1}{v}=-\frac{1}{15}+\frac{1}{30}
\]
\[
\frac{1}{v}=\frac{-2+1}{30}=-\frac{1}{30}
\]
\[
v=-30\,cm
\]
The negative value places the real image in front of the mirror.
\( \textbf{Final answer:} \) \(-30\,cm\), matching the object-at-\(C\) case.
103. In the situation \(f=-15\,cm\), \(u=-30\,cm\), and \(v=-30\,cm\) for a concave mirror, the magnification is
ⓐ. \(-1\)
ⓑ. \(+1\)
ⓒ. \(+2\)
ⓓ. \(-2\)
Correct Answer: \(-1\)
Explanation: \( \textbf{Known values:} \)
Object distance is \(u=-30\,cm\).
Image distance is \(v=-30\,cm\).
Mirror magnification is
\[
m=-\frac{v}{u}
\]
Substituting,
\[
m=-\frac{-30}{-30}
\]
\[
m=-1
\]
The magnitude \(|m|=1\) means the image size equals the object size.
The negative sign means the image is inverted.
\( \textbf{Final answer:} \) \(m=-1\), so the image is same-sized and inverted.
104. A concave mirror has \(f=-20\,cm\). A \(3\,cm\) tall object is placed \(30\,cm\) in front of it. The image height is
ⓐ. \(+6\,cm\)
ⓑ. \(-2\,cm\)
ⓒ. \(+2\,cm\)
ⓓ. \(-6\,cm\)
Correct Answer: \(-6\,cm\)
Explanation: \( \textbf{Given data:} \)
Focal length is \(f=-20\,cm\).
Object distance is \(u=-30\,cm\).
Object height is \(h_o=+3\,cm\).
First find \(v\) using
\[
\frac{1}{f}=\frac{1}{v}+\frac{1}{u}
\]
\[
\frac{1}{-20}=\frac{1}{v}+\frac{1}{-30}
\]
\[
\frac{1}{v}=-\frac{1}{20}+\frac{1}{30}
\]
\[
\frac{1}{v}=\frac{-3+2}{60}=-\frac{1}{60}
\]
\[
v=-60\,cm
\]
Now use mirror magnification:
\[
m=-\frac{v}{u}=-\frac{-60}{-30}=-2
\]
Image height is
\[
h_i=mh_o=(-2)(3\,cm)=-6\,cm
\]
\( \textbf{Final answer:} \) \(-6\,cm\); the negative height shows that the enlarged image is inverted.
105. A convex mirror of focal length \(+15\,cm\) forms an image of an object placed \(30\,cm\) in front of it. If the object height is \(6\,cm\), the image height is
ⓐ. \(+2\,cm\)
ⓑ. \(-2\,cm\)
ⓒ. \(+12\,cm\)
ⓓ. \(-12\,cm\)
Correct Answer: \(+2\,cm\)
Explanation: \( \textbf{Given data:} \)
Convex mirror focal length is \(f=+15\,cm\).
Object distance is \(u=-30\,cm\).
Object height is \(h_o=+6\,cm\).
Apply the mirror formula:
\[
\frac{1}{15}=\frac{1}{v}+\frac{1}{-30}
\]
\[
\frac{1}{v}=\frac{1}{15}+\frac{1}{30}
\]
\[
\frac{1}{v}=\frac{2+1}{30}=\frac{3}{30}=\frac{1}{10}
\]
\[
v=+10\,cm
\]
Magnification is
\[
m=-\frac{v}{u}=-\frac{+10}{-30}=+\frac{1}{3}
\]
Therefore,
\[
h_i=mh_o=\frac{1}{3}\times6\,cm=2\,cm
\]
\( \textbf{Final answer:} \) \(+2\,cm\), so the image is erect and diminished.
106. A mirror gives \(m=-3\) for a real object. This means the image is
ⓐ. erect and three times as tall as the object
ⓑ. inverted and three times as tall as the object
ⓒ. erect and one-third as tall as the object
ⓓ. inverted and one-third as tall as the object
Correct Answer: inverted and three times as tall as the object
Explanation: Magnification \(m\) contains both sign and magnitude information. A negative value of \(m\) means the image is inverted relative to the object. The magnitude \(|m|=3\) means the image height is three times the object height. Therefore \(m=-3\) describes an inverted enlarged image. The sign does not mean the image has negative physical size; it shows orientation with respect to the principal axis.
107. A mirror produces a virtual erect image of height \(4\,cm\) from an object of height \(2\,cm\). The magnification is
ⓐ. \(-2\)
ⓑ. \(+\frac{1}{2}\)
ⓒ. \(-\frac{1}{2}\)
ⓓ. \(+2\)
Correct Answer: \(+2\)
Explanation: \( \textbf{Given data:} \)
Object height is \(h_o=+2\,cm\).
The image is erect, so image height is positive: \(h_i=+4\,cm\).
Magnification is
\[
m=\frac{h_i}{h_o}
\]
Substituting,
\[
m=\frac{+4}{+2}
\]
\[
m=+2
\]
The positive sign agrees with an erect image.
The magnitude \(2\) shows that the image is twice as tall as the object.
\( \textbf{Final answer:} \) \(+2\).
108. A graph is plotted for a spherical mirror with \(y=\frac{1}{v}\) on the vertical axis and \(x=\frac{1}{u}\) on the horizontal axis. From the mirror formula, the graph should have
ⓐ. slope \(+1\) and vertical intercept \(f\)
ⓑ. slope \(0\) and vertical intercept \(u\)
ⓒ. slope \(\frac{1}{f}\) and vertical intercept \(-1\)
ⓓ. slope \(-1\) and vertical intercept \(\frac{1}{f}\)
Correct Answer: slope \(-1\) and vertical intercept \(\frac{1}{f}\)
Explanation: \( \textbf{Starting relation:} \)
The mirror formula is
\[
\frac{1}{f}=\frac{1}{v}+\frac{1}{u}
\]
Rearranging for \(\frac{1}{v}\),
\[
\frac{1}{v}=\frac{1}{f}-\frac{1}{u}
\]
Let
\[
y=\frac{1}{v}
\]
and
\[
x=\frac{1}{u}
\]
Then the equation becomes
\[
y=\frac{1}{f}-x
\]
This is a straight-line form with slope \(-1\).
The vertical intercept is \(\frac{1}{f}\).
\( \textbf{Final answer:} \) slope \(-1\) and vertical intercept \(\frac{1}{f}\).
109. A dentist uses a concave mirror of focal length \(15\,cm\) to see an enlarged erect image of a tooth. If the tooth is \(10\,cm\) in front of the mirror, the image distance and magnification are
ⓐ. \(v=-30\,cm,\ m=-3\)
ⓑ. \(v=+30\,cm,\ m=+3\)
ⓒ. \(v=+6\,cm,\ m=+0.6\)
ⓓ. \(v=-6\,cm,\ m=-0.6\)
Correct Answer: \(v=+30\,cm,\ m=+3\)
Explanation: \( \textbf{Given data:} \)
The mirror is concave, so \(f=-15\,cm\).
The tooth is in front of the mirror, so \(u=-10\,cm\).
Use the mirror formula:
\[
\frac{1}{f}=\frac{1}{v}+\frac{1}{u}
\]
\[
\frac{1}{-15}=\frac{1}{v}+\frac{1}{-10}
\]
\[
\frac{1}{v}=-\frac{1}{15}+\frac{1}{10}
\]
\[
\frac{1}{v}=\frac{-2+3}{30}=\frac{1}{30}
\]
\[
v=+30\,cm
\]
The positive image distance places the virtual image behind the mirror.
Magnification is
\[
m=-\frac{v}{u}=-\frac{+30}{-10}=+3
\]
\( \textbf{Final answer:} \) \(v=+30\,cm\) and \(m=+3\), giving an erect enlarged virtual image.
110. A student solves a concave-mirror problem and obtains \(v=+24\,cm\) for a real object placed in front of the mirror. The physical interpretation is that the image is
ⓐ. behind the mirror and virtual
ⓑ. in front of the mirror and real
ⓒ. at the pole and real
ⓓ. below the principal axis only
Correct Answer: behind the mirror and virtual
Explanation: In the usual mirror sign convention, the pole is the origin and incident light is taken as the positive direction. For a mirror facing the object, the region behind the mirror is on the positive side. Therefore \(v=+24\,cm\) places the image behind the mirror. Since reflected rays do not actually pass behind the mirror and meet there, this image is virtual. The sign of \(v\) gives image position along the principal axis, while image height or magnification is needed to describe orientation.
111. A concave mirror forms a real image \(45\,cm\) in front of it when an object is placed \(30\,cm\) in front of the mirror. The focal length of the mirror is
ⓐ. \(-18\,cm\)
ⓑ. \(+18\,cm\)
ⓒ. \(-75\,cm\)
ⓓ. \(+75\,cm\)
Correct Answer: \(-18\,cm\)
Explanation: \( \textbf{Given data:} \)
A real object in front of the mirror has \(u=-30\,cm\).
A real image in front of a concave mirror has \(v=-45\,cm\).
The mirror formula is
\[
\frac{1}{f}=\frac{1}{v}+\frac{1}{u}
\]
Substituting the signed values,
\[
\frac{1}{f}=\frac{1}{-45}+\frac{1}{-30}
\]
\[
\frac{1}{f}=-\frac{2}{90}-\frac{3}{90}=-\frac{5}{90}
\]
\[
\frac{1}{f}=-\frac{1}{18}
\]
\[
f=-18\,cm
\]
The negative sign agrees with a concave mirror whose focus lies in front of the mirror.
\( \textbf{Final answer:} \) \(-18\,cm\).
112. A spherical mirror gives \(u=-24\,cm\) and \(v=+8\,cm\) for a real object. The mirror type and image nature are
ⓐ. concave mirror, real image
ⓑ. convex mirror, virtual image
ⓒ. concave mirror, virtual image
ⓓ. convex mirror, real image
Correct Answer: convex mirror, virtual image
Explanation: \( \textbf{Given signs:} \)
The object distance \(u=-24\,cm\) shows a real object in front of the mirror.
The image distance \(v=+8\,cm\) places the image behind the mirror.
An image behind a mirror is virtual because reflected rays do not actually pass behind the mirror and meet there.
Find the focal length using
\[
\frac{1}{f}=\frac{1}{v}+\frac{1}{u}
\]
\[
\frac{1}{f}=\frac{1}{8}+\frac{1}{-24}
\]
\[
\frac{1}{f}=\frac{3-1}{24}=\frac{2}{24}=\frac{1}{12}
\]
\[
f=+12\,cm
\]
A positive focal length identifies a convex mirror in the usual mirror convention.
\( \textbf{Final answer:} \) convex mirror, virtual image.
113. A mirror forms an image twice the size of the object and inverted. If the object is \(18\,cm\) in front of the mirror, the image distance is
ⓐ. \(+36\,cm\)
ⓑ. \(-9\,cm\)
ⓒ. \(-36\,cm\)
ⓓ. \(+9\,cm\)
Correct Answer: \(-36\,cm\)
Explanation: \( \textbf{Given data:} \)
Object distance is \(u=-18\,cm\).
The image is inverted and twice the size of the object, so \(m=-2\).
For a mirror,
\[
m=-\frac{v}{u}
\]
Substitute \(m=-2\) and \(u=-18\,cm\):
\[
-2=-\frac{v}{-18}
\]
\[
-2=\frac{v}{18}
\]
\[
v=-36\,cm
\]
The negative value places the image in front of the mirror.
This matches an actual convergence of reflected rays, so the image is real and inverted.
\( \textbf{Final answer:} \) \(-36\,cm\).
114. A concave mirror of radius of curvature \(50\,cm\) is used to form a real image of an object placed \(75\,cm\) in front of it. The magnification is
ⓐ. \(+\frac{1}{2}\)
ⓑ. \(-2\)
ⓒ. \(-\frac{1}{2}\)
ⓓ. \(+2\)
Correct Answer: \(-\frac{1}{2}\)
Explanation: \( \textbf{Given data:} \)
Concave mirror radius is \(R=-50\,cm\), so
\[
f=\frac{R}{2}=-25\,cm
\]
Object distance is \(u=-75\,cm\).
Use the mirror formula:
\[
\frac{1}{f}=\frac{1}{v}+\frac{1}{u}
\]
\[
\frac{1}{-25}=\frac{1}{v}+\frac{1}{-75}
\]
\[
\frac{1}{v}=-\frac{1}{25}+\frac{1}{75}
\]
\[
\frac{1}{v}=\frac{-3+1}{75}=-\frac{2}{75}
\]
\[
v=-37.5\,cm
\]
Now use
\[
m=-\frac{v}{u}
\]
\[
m=-\frac{-37.5}{-75}=-\frac{1}{2}
\]
The negative sign shows inversion, and the magnitude \(\frac{1}{2}\) shows diminution.
\( \textbf{Final answer:} \) \(-\frac{1}{2}\).
115. A data sheet for a mirror experiment is given below:
| Row | Given result | Reasonable interpretation |
|---|
| P | \(v\lt 0\) for a real object in front of a concave mirror | real image in front of mirror |
| Q | \(v\gt 0\) for a real object in front of a convex mirror | virtual image behind mirror |
| R | \(m\lt 0\) | erect image |
| S | \(|m|\gt1\) | image larger than object |
The acceptable rows are
ⓐ. P and R only
ⓑ. P, Q, and S only
ⓒ. Q, R, and S only
ⓓ. P, Q, R, and S
Correct Answer: P, Q, and S only
Explanation: Row P is acceptable because a negative \(v\) in the usual mirror convention places the image in front of the mirror, where a concave mirror can form a real image. Row Q is acceptable because a convex mirror forms a virtual image behind the mirror, giving \(v\gt 0\). Row S is acceptable because magnification magnitude greater than \(1\) means the image height is greater than the object height. Row R is not acceptable because \(m\lt 0\) represents an inverted image, not an erect image. The sign of \(m\) carries orientation, while the magnitude of \(m\) carries size comparison.
116. A mirror has \(f=-12\,cm\), and a virtual image is formed \(24\,cm\) behind the mirror. The object distance is
ⓐ. \(+8\,cm\)
ⓑ. \(-24\,cm\)
ⓒ. \(-8\,cm\)
ⓓ. \(+24\,cm\)
Correct Answer: \(-8\,cm\)
Explanation: \( \textbf{Given data:} \)
The mirror has \(f=-12\,cm\), so it is concave.
A virtual image behind the mirror has \(v=+24\,cm\).
The mirror formula is
\[
\frac{1}{f}=\frac{1}{v}+\frac{1}{u}
\]
Substitute the known values:
\[
\frac{1}{-12}=\frac{1}{24}+\frac{1}{u}
\]
\[
\frac{1}{u}=-\frac{1}{12}-\frac{1}{24}
\]
\[
\frac{1}{u}=-\frac{2}{24}-\frac{1}{24}=-\frac{3}{24}
\]
\[
\frac{1}{u}=-\frac{1}{8}
\]
\[
u=-8\,cm
\]
The object lies between \(F\) and \(P\), which is the concave-mirror region for a virtual enlarged image.
\( \textbf{Final answer:} \) \(-8\,cm\).
117. A mirror calculation gives \(m=+0.6\) for a real object. This result is most consistent with
ⓐ. a real inverted diminished image
ⓑ. a virtual inverted enlarged image
ⓒ. a real erect enlarged image
ⓓ. a virtual erect diminished image
Correct Answer: a virtual erect diminished image
Explanation: A positive magnification means the image is erect relative to the object. The magnitude \(|m|=0.6\) is less than \(1\), so the image is diminished. For a single spherical mirror with a real object, an erect image is virtual in the standard cases. A convex mirror naturally gives this result for all real object positions. A concave mirror can give an erect image too, but that image is enlarged when the object is between \(F\) and \(P\), not diminished.
118. In a graph for a mirror experiment, \(m\) is plotted against \(v\) while the object distance \(u\) is kept fixed at \(-20\,cm\). The relation between \(m\) and \(v\) is
ⓐ. a straight line with slope \(-20\,cm\)
ⓑ. a curve because \(m=-uv\)
ⓒ. a straight line with slope \(+\frac{1}{20\,cm}\)
ⓓ. a horizontal line because \(m\) cannot change with \(v\)
Correct Answer: a straight line with slope \(+\frac{1}{20\,cm}\)
Explanation: \( \textbf{Starting relation:} \)
Mirror magnification is
\[
m=-\frac{v}{u}
\]
The object distance is fixed at
\[
u=-20\,cm
\]
Substitute this value:
\[
m=-\frac{v}{-20\,cm}
\]
\[
m=\frac{v}{20\,cm}
\]
This is a straight-line relation between \(m\) and \(v\).
The slope of the \(m\)-versus-\(v\) graph is
\[
\frac{1}{20\,cm}
\]
The sign comes from using \(u=-20\,cm\), not from treating all distances as positive numbers.
\( \textbf{Final answer:} \) a straight line with slope \(+\frac{1}{20\,cm}\).
119. When a light ray passes obliquely from air into glass, it bends toward the normal mainly because
ⓐ. its frequency becomes zero in glass
ⓑ. the normal attracts the ray like a force
ⓒ. the glass surface reflects all the light
ⓓ. its speed decreases in glass
Correct Answer: its speed decreases in glass
Explanation: Refraction occurs when light passes from one transparent medium into another and its speed changes. Glass is optically denser than air, so light travels slower in glass than in air. For oblique incidence from air into glass, this speed decrease makes the ray bend toward the normal. The frequency of light remains unchanged at the boundary, while wavelength changes along with speed. The normal is a reference line for measuring angles, not a physical line that pulls the ray.
120. A ray goes from glass into air at an oblique angle. The refracted ray bends
ⓐ. toward the normal because air is optically denser
ⓑ. along the normal for every angle of incidence
ⓒ. away from the normal because air is optically rarer
ⓓ. back into the glass without any refracted ray for every angle
Correct Answer: away from the normal because air is optically rarer
Explanation: Glass has a higher refractive index than air, so glass is optically denser and air is optically rarer. When light travels from a denser medium to a rarer medium at oblique incidence, it speeds up and bends away from the normal. This does not mean the ray always disappears at the boundary. Total internal reflection is possible only when additional angle conditions are satisfied. For ordinary refraction below the critical condition, a refracted ray appears in air and makes a larger angle with the normal.
