Numerical Problems Based on Mirror Formula for Class 12 Physics

Answer: $v = +8\text{ cm}$, Size $= 4\text{ cm}$, Virtual & Erect. Size decreases as object moves away.

Explanation: Given $h = +5\text{ cm}$, $u = -10\text{ cm}$, and for a convex mirror, $f = +40\text{ cm}$.

So, $v = +8\text{ cm}$. The positive sign indicates the image is virtual and forms behind the mirror.

As the needle is moved further away ($u \to -\infty$), the image moves towards the focus ($v \to 40\text{ cm}$) and its size diminishes progressively until it becomes a point-sized image.

Answer: (i) $v = -30\text{ cm}, m = -3$ (Real, inverted) (ii) $v = +15\text{ cm}, m = +3$ (Virtual, erect)

Explanation: For a concave mirror, $R = -15\text{ cm}$, so $f = -7.5\text{ cm}$.

Case (i): $u = -10\text{ cm}$

So, $v = -30\text{ cm}$. Magnification $m = -\frac{-30}{-10} = -3$. The image is real, inverted, and magnified.

Case (ii): $u = -5\text{ cm}$

So, $v = +15\text{ cm}$. Magnification $m = -\frac{15}{-5} = +3$. The image is virtual, erect, and magnified.

Answer: 21 cm from the wall

Explanation: The wall acts as a screen where the real image is formed. Thus, the image distance from the mirror is $v = -35\text{ cm}$. Given $f = -10\text{ cm}$. Let’s find object distance $u$:

The object is placed 14 cm in front of the mirror. The total distance to the wall is 35 cm. Therefore, the distance of the object from the wall is $35 – 14 = 21\text{ cm}$.

Answer: 1.5

Explanation: For a real image to just touch the object, one end of the rod must be located at the center of curvature (where $u = -2f$). At this point, the image forms exactly at the same place ($v = -2f$).

Since the image is elongated, the rod must stretch from $C$ towards the focus $F$. Thus, the other end of the rod is at $u_2 = -(2f – f/3) = -5f/3$.

We find the image position $v_2$ for this second end:

The length of the image is $|v_2 – v_1| = |-2.5f – (-2f)| = 0.5f = f/2$.

The longitudinal magnification is the ratio of image length to object length:

Answer: 20 cm and 10 cm

Explanation: Given $f = -15\text{ cm}$. A concave mirror can form both real ($m = -3$) and virtual ($m = +3$) magnified images.

Case 1: Real Image ($m = -3$)
$m = -v/u \implies -3 = -v/u \implies v = 3u$.

Case 2: Virtual Image ($m = +3$)
$m = -v/u \implies 3 = -v/u \implies v = -3u$.

Answer: The image shifts 13.5 cm away from the mirror.

Explanation: Given $R = -30\text{ cm} \implies f = -15\text{ cm}$.

Initial state ($u_1 = -40\text{ cm}$):

Final state: The object is shifted towards the mirror by 15 cm, so $u_2 = -40 + 15 = -25\text{ cm}$.

The image shifts from 24 cm to 37.5 cm. Total shift = $37.5 – 24 = 13.5\text{ cm}$ away from the mirror.

Answer: $f = -6\text{ cm}$ (if both real) or $f = -10.5\text{ cm}$ (if one real, one virtual)

Explanation: Magnification is given by $m = \frac{f}{f – u}$. Let’s solve assuming the problem implies both images are real ($m_1$ and $m_2$ have the same sign).

Condition: $m_2 = 3m_1$

Note: If the second image became virtual, the relation would be $m_2 = -3m_1$, which yields $f = -10.5\text{ cm}$. Standard exams usually accept the real-real case unless specified.

Answer: Object distance = 15 cm in front, Image distance = 30 cm in front

Explanation: $R = -20\text{ cm} \implies f = -10\text{ cm}$. For a real, magnified image, $m = -2$.

Applying the mirror formula:

The location of the image is $v = 2(-15) = -30\text{ cm}$.

Answer: 20 cm in front of the mirror

Explanation: Given $f = -10\text{ cm}$, $h_A = 4 h_B$, and $|h’_A| = |h’_B|$.

First, find the magnification of object A ($u_A = -50\text{ cm}$):

Because $|h’_A| = |m_A| h_A$ and $|h’_B| = |m_B| h_B$, setting them equal gives:

Since object B must form an image of magnification magnitude 1, it must be placed at the center of curvature (where $m = -1$).

Answer: $R = 36\text{ cm}$

Explanation: The plane mirror is 24 cm from the object, so its virtual image forms 24 cm behind the plane mirror. Total distance from the object to this image is $24 + 24 = 48\text{ cm}$.

Since the convex mirror is 36 cm away from the object, the distance from the convex mirror to the coincident image is $48 – 36 = 12\text{ cm}$. Therefore, for the convex mirror, $v = +12\text{ cm}$ and $u = -36\text{ cm}$.

Focal length $f = +18\text{ cm}$. The radius of curvature is $R = 2f = 36\text{ cm}$.

Answer: $4.0\text{ cm}^2$

Explanation: First, find the image distance ($v$) for $u = -25\text{ cm}$ and $f = -10\text{ cm}$.

The transverse (linear) magnification is:

The length of the sides of the square image will be $|m| \times \text{side} = \frac{2}{3} \times 3.0 = 2.0\text{ cm}$.

Therefore, the area of the image is $(2.0)^2 = 4.0\text{ cm}^2$.

Answer: (a) $1/320\text{ m/s}$ (b) $1/180\text{ m/s}$ (c) $1/80\text{ m/s}$ (d) $1/20\text{ m/s}$

Explanation: Rear-view mirror is convex, so $f = R/2 = +1\text{ m}$. Jogger’s speed $v_o = 5\text{ m/s}$. The speed of the image $v_i$ is given by $v_i = m^2 \cdot v_o$, where magnification $m = \frac{f}{f – u}$.

(a) $u = -39\text{ m}$:

(b) $u = -29\text{ m}$:

(c) $u = -19\text{ m}$:

(d) $u = -9\text{ m}$: