Question

A 5 cm long pencil is placed along the principal axis of a concave mirror of focal length 20 cm such that its nearest end is at a distance of 25 cm from the mirror. Calculate the length of the image of the pencil.

Hint:

For objects placed completely horizontally along the principal axis (longitudinal magnification), never just use the transverse magnification formula $m = -v/u$ directly on the length. Always rigorously find the two end points separately.

Answer 1

Concept:
• When an extended object (like a long pencil) is placed perfectly horizontally along the principal axis of a spherical mirror, its final resulting image will inherently possess a specific measurable length.

• This longitudinal image length is meticulously found by treating the two extreme physical ends of the object as two entirely separate point objects.

• We robustly apply the standard spherical mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ twice, independently, to locate the precise spatial positions of the images of both ends.

• The absolute final length of the image is simply the mathematical magnitude of the spatial difference between these two computed image positions.

Step 1: Identify the mirror parameters using Cartesian sign convention
The optical device is explicitly a concave mirror. Therefore, its primary focal point lies in front of it, meaning its focal length is strictly negative.
Focal length, $f = -20 \text{ cm}$.
The entire object is placed physically in front of the reflecting surface, so all object distances are strictly negative.

Step 2: Calculate the image position for the nearest end of the pencil
The nearest end of the 5 cm pencil is stated to be exactly 25 cm from the mirror's pole.
So, $u_1 = -25 \text{ cm}$.
Apply the mirror formula:
\[ \frac{1}{v_1} + \frac{1}{u_1} = \frac{1}{f} \]
\[ \frac{1}{v_1} + \frac{1}{-25} = \frac{1}{-20} \]
Rearrange to isolate the unknown image distance term:
\[ \frac{1}{v_1} = \frac{1}{25} - \frac{1}{20} \]
Find the lowest common denominator (100) to subtract the fractions smoothly:
\[ \frac{1}{v_1} = \frac{4 - 5}{100} = \frac{-1}{100} \]
Taking the reciprocal solidly yields the position of the first end's image:
\[ v_1 = -100 \text{ cm} \]

Step 3: Calculate the image position for the farthest end of the pencil
Since the pencil is exactly 5 cm long and lies completely along the axis starting from 25 cm, its other, farther end must logically be at a distance of $25 + 5 = 30 \text{ cm}$.
So, $u_2 = -30 \text{ cm}$.
Apply the mirror formula again independently:
\[ \frac{1}{v_2} + \frac{1}{-30} = \frac{1}{-20} \]
Rearrange to isolate the unknown image distance term:
\[ \frac{1}{v_2} = \frac{1}{30} - \frac{1}{20} \]
Find the lowest common denominator (60) to subtract the fractions smoothly:
\[ \frac{1}{v_2} = \frac{2 - 3}{60} = \frac{-1}{60} \]
Taking the reciprocal solidly yields the position of the second end's image:
\[ v_2 = -60 \text{ cm} \]

Step 4: Determine the final longitudinal length of the image
Both ends of the pencil generate real images positioned at 100 cm and 60 cm in front of the mirror respectively.
The total physical length of this extended image is simply the absolute difference between these two distinct coordinates:
\[ \text{Length of image} = |v_1 - v_2| \]
\[ \text{Length of image} = |-100 - (-60)| = |-100 + 60| = |-40| \]
\[ \text{Length of image} = 40 \text{ cm} \]

Step 5: Conclusion
The calculated length of the pencil's image is dramatically stretched out to 40 cm. Since both image coordinates are negative, it forms a completely real image lying entirely in front of the mirror.