0.5 m or 50 cm.

Explanation: For a spherical mirror, the focal length is half of the radius of curvature ($f = R/2$). Since $R = 1$ m, $f = 1/2 = 0.5$ m.

1 m or 100 cm.

Explanation: Radius of curvature $R = 2f$. Given $f = 50$ cm, $R = 2 \times 50 = 100$ cm.

12.5 cm

Explanation: $f = R/2$. Given magnitude $R = 25$ cm, the focal length magnitude is $25 / 2 = 12.5$ cm. (By sign convention, $f = -12.5$ cm for a concave mirror).

-5

Explanation: Magnification $m = \frac{h’}{h}$. Assuming a standard real, inverted image for a concave mirror, $h’ = -10$ cm and $h = 2$ cm. $m = \frac{-10}{2} = -5$.

10 cm

Explanation: Magnification $m = \frac{h’}{h}$. Given $m = 10$ and $h = 1$ cm. Therefore, image height $h’ = m \times h = 10 \times 1 = 10$ cm. (Note: If the image is real and inverted, $m = -10$ and $h’ = -10$ cm. Size is 10 cm either way.)

Given: $u = -25$ cm, $f = -15$ cm, $h = +4$ cm.

Position: $v = -37.5$ cm (in front of the mirror).

Magnification: $m = -\frac{v}{u} = -\frac{-37.5}{-25} = -1.5$.

Height: $h’ = m \times h = -1.5 \times 4 = -6$ cm.

Nature: Real, inverted, and magnified.

Given: $h = +1$ cm, $h’ = -4$ cm (real image is inverted), $u = -20$ cm.

Magnification $m = \frac{h’}{h} = -4$. Also, $m = -\frac{v}{u}$.

Image distance is 80 cm in front of the mirror.

Focal length: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{-80} + \frac{1}{-20} = \frac{-1 – 4}{80} = -\frac{5}{80}$

$f = -16$ cm.

Given: $h = 4.5$ cm, $u = -12$ cm, $f = +15$ cm (convex mirror).

Magnification: $m = -\frac{v}{u} = -\frac{60/9}{-12} = +\frac{5}{9} \approx +0.55$.

As the needle moves farther: The virtual image moves towards the principal focus ($v \to 15$ cm) and its size continuously decreases.

Given: Diverging (convex) mirror, $f = +20$ cm, $u = -25$ cm, $h = 2.5$ cm.

Position: $v = \frac{100}{9} \approx +11.11$ cm (behind mirror).

Magnification: $m = -\frac{v}{u} = -\frac{100/9}{-25} = +\frac{4}{9}$.

Size: $h’ = m \times h = \frac{4}{9} \times 2.5 \approx 1.11$ cm.

Nature: Virtual, erect, and diminished.

Given: Convex mirror ($f = +20$ cm). Image is always virtual and erect, so $m = +1/4$.

$m = -\frac{v}{u} \implies \frac{1}{4} = -\frac{v}{u} \implies v = -\frac{u}{4}$.

Solving for $u$: $u = -60$ cm. Distance is 60 cm.

Given: $u = -15$ cm, $f = -10$ cm, $h = 1$ cm.

Magnification $m = -\frac{v}{u} = -\frac{-30}{-15} = -2$.

Size $h’ = m \times h = -2 \times 1 = -2$ cm.

Position: 30 cm in front of mirror. Nature: Real and inverted. Size: 2 cm.

Given: $h = 2$ cm, $u = -16$ cm, $h’ = -3$ cm (inverted).

Magnification $m = \frac{h’}{h} = -\frac{3}{2} = -1.5$.

Position: $m = -\frac{v}{u} \implies -1.5 = -\frac{v}{-16} \implies v = -24$ cm.

Focal Length: $\frac{1}{f} = \frac{1}{-24} + \frac{1}{-16} = \frac{-2 – 3}{48} = -\frac{5}{48}$

$f = -9.6$ cm.

Given: Concave mirror, $R = -36$ cm $\implies f = -18$ cm. Erect image means virtual, so $m = +3$.

$m = -\frac{v}{u} = 3 \implies v = -3u$.

$-3u = 36 \implies u = -12$ cm. Position of object is 12 cm in front of the mirror.

Given: $u = -12$ cm, $f = +30$ cm, $h = 2.5$ cm.

Location: $8.57$ cm behind the mirror.

Magnification: $m = -\frac{v}{u} = -\frac{60/7}{-12} = +\frac{5}{7} \approx 0.71$.

Given: $f = -20$ cm. A concave mirror can form both real ($m=-3$) and virtual ($m=+3$) magnified images.

Case 1 (Real Image): $m = -3 \implies v = 3u$.
$-\frac{1}{20} = \frac{1}{3u} + \frac{1}{u} = \frac{4}{3u} \implies u = -26.67$ cm.

Case 2 (Virtual Image): $m = +3 \implies v = -3u$.
$-\frac{1}{20} = -\frac{1}{3u} + \frac{1}{u} = \frac{2}{3u} \implies u = -13.33$ cm.

In a convex mirror, a ray parallel to the principal axis appears to diverge from the principal focus ($F$). A ray directed towards the center of curvature ($C$) reflects back along its own path. These two reflected rays diverge, but when extended backwards, they intersect behind the mirror between the pole ($P$) and $F$, forming a virtual, erect, and diminished image.

Given: $h = 2.5$ mm, $h’ = -10$ mm (real image is inverted), $u = -5$ cm.

Magnification $m = \frac{-10}{2.5} = -4$.

Position: $m = -\frac{v}{u} \implies -4 = -\frac{v}{-5} \implies v = -20$ cm.

Focal Length: $\frac{1}{f} = \frac{1}{-20} + \frac{1}{-5} = \frac{-1 – 4}{20} = -\frac{5}{20} \implies f = -4$ cm.

Given: $R = +40$ cm $\implies f = +20$ cm.

When an object is placed at a very large distance ($u \to -\infty$), parallel rays strike the mirror and appear to converge at the principal focus.

Therefore, the image is formed at $v = f = +20$ cm. Distance is 20 cm behind the mirror.

Given: $u = -15$ cm, $R = +90$ cm $\implies f = +45$ cm.

Position: $v = \frac{45}{4} = 11.25$ cm (behind mirror).

Magnification: $m = -\frac{v}{u} = -\frac{11.25}{-15} = +0.75$.

Given: $f = +30$ cm, $m = +1/4$ (convex mirrors form virtual, erect images).

$v = -u \times m = -\frac{u}{4}$.

Solving for $u$: $u = -90$ cm. Distance is 90 cm.

Step 1: Find focal length. $u_1 = -60$ cm, $m_1 = 1/2 \implies v_1 = 30$ cm.
$\frac{1}{f} = \frac{1}{30} – \frac{1}{60} = \frac{1}{60} \implies f = +60$ cm.

Step 2: Find new object distance. $m_2 = 1/3 \implies v_2 = -u_2/3$.
$\frac{1}{60} = -\frac{3}{u_2} + \frac{1}{u_2} = -\frac{2}{u_2} \implies u_2 = -120$ cm.

The object should be placed at 120 cm.

Given: $u = -18$ cm. Image to the right means $v = +4$ cm.

Focal Length: $f = \frac{36}{7} \approx +5.14$ cm.

Because $f$ is positive, it is a Convex Mirror. The image is Virtual, erect, and diminished.

Radius of curvature $R = 2f = \frac{72}{7} \approx 10.28$ cm.

Given: $R = +3$ m $\implies f = +1.5$ m. $u = -5$ m.

Position: $v = \frac{15}{13} \approx +1.15$ m (behind mirror).

Magnification: $m = -\frac{v}{u} = -\frac{1.15}{-5} \approx +0.23$.

Nature: Virtual, erect, and diminished.

Given: Diverging (convex) mirror, $f = +25$ cm, $u = -50$ cm, $h = 3$ cm.

Magnification: $m = -\frac{v}{u} = -\frac{50/3}{-50} = +\frac{1}{3}$.

Size: $h’ = m \times h = \frac{1}{3} \times 3 = 1$ cm.

Nature: Virtual, erect, and diminished.

Given: Converging (concave) mirror, $f = -20$ cm, $m = \pm 2$.

Case 1 (Real): $m = -2 \implies v = 2u$.
$-\frac{1}{20} = \frac{1}{2u} + \frac{1}{u} = \frac{3}{2u} \implies 2u = -60 \implies u = -30$ cm.

Case 2 (Virtual): $m = +2 \implies v = -2u$.
$-\frac{1}{20} = -\frac{1}{2u} + \frac{1}{u} = \frac{1}{2u} \implies 2u = -20 \implies u = -10$ cm.

Given: $f = +15$ cm, $m = +1/3 \implies v = -u/3$.

The object is 30 cm in front of the mirror. The principal focus is 15 cm behind the mirror. Therefore, the absolute distance between the object and the focus is $30 + 15 = \mathbf{45}$ cm.

Given: $f = 12.5$ cm. The radius of curvature is $R = 2f = 25$ cm.

(i) From the pole: The center of curvature is at a distance of $R = \mathbf{25}$ cm.

(ii) From the focus: The focus lies exactly halfway between the pole and the center of curvature, so the distance is $25 – 12.5 = \mathbf{12.5}$ cm.

Given: $u = -5$ cm. Real image is inverted, so $m = -4$.

$v = -m \times u = -(-4) \times (-5) = -20$ cm.

Convex Mirror Image: $u = -30$, $f = +15$. $\frac{1}{v} = \frac{1}{15} – (-\frac{1}{30}) = \frac{3}{30} \implies v = +10$ cm. The image is 10 cm behind the convex mirror.

Total distance between object and convex mirror’s image is $30 + 10 = 40$ cm.

For the plane mirror image to coincide, the plane mirror must be placed exactly halfway between the object and the final image. Half of 40 cm is 20 cm from the object.

Distance of plane mirror from the convex mirror = $30 – 20 = \mathbf{10}$ cm in front of the convex mirror.

As the object moves from C to P, the three key stages to illustrate with ray diagrams are:

• Object between C and F: The image is formed beyond C, and is real, inverted, and magnified.
• Object at F: The reflected rays become parallel, forming a highly magnified real image at infinity.
• Object between F and P: The reflected rays diverge, extending backward to form a virtual, erect, and magnified image behind the mirror.