Q.A small candle, $2.5\ \text{cm}$ in size is placed at $27\ \text{cm}$ in front of a concave mirror of radius of curvature $36\ \text{cm}$. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?

Concept: Reflection by a spherical mirror — use the mirror formula and magnification relation.

Step 1 — Find focal length.

For a concave mirror, $f = \frac{R}{2} = \frac{36}{2} = 18\ \text{cm}$.

Object distance $u = -27\ \text{cm}$ (sign convention: object in front).

Step 2 — Apply mirror formula.

Thus $v = -54\ \text{cm}$. The negative sign means the image is real and on the same side as the object.

Step 3 — Magnification and image size.

Image height $h_i = m \times h_o = -2 \times 2.5 = -5\ \text{cm}$.

Negative $m$ means inverted; $|m|>1$ means enlarged.

Step 4 — Effect of moving candle closer.

If the candle is moved closer (i.e., $|u|$ decreases), $v$ becomes more negative (image moves farther behind the mirror). Since the image is real, the screen must be moved away from the mirror to keep the image sharp.

1. Understanding the physics: Why the mirror formula works

A concave mirror converges light. When an object is placed beyond the centre of curvature ($C$), the image forms between $C$ and $F$ — real and inverted. When the object is between $F$ and the pole, the image is virtual and erect. The mirror formula ties object distance $u$, image distance $v$, and focal length $f$:

Sign convention (Cartesian): distances measured from the pole. For a concave mirror, $f$ is negative, $u$ is negative (object in front), and $v$ is negative for a real image (in front of the mirror).

2. Step-by-step solution

Step 1: Find the focal length.

Radius of curvature $R = 36\ \text{cm}$. For any spherical mirror, $f = R/2$.

Since it's concave, $f = -18\ \text{cm}$.

Step 2: Write the object distance.

Object is placed $27\ \text{cm}$ in front of the mirror.

Step 3: Apply the mirror formula.

Find a common denominator (LCM = 54):

Thus:

Step 4: Interpret $v$.

The negative sign means the image is formed in front of the mirror — real and inverted. The screen must be placed $54\ \text{cm}$ from the mirror on the same side as the object.

Step 5: Find the magnification and image size.

Magnification $m = -\frac{v}{u}$:

The negative sign indicates inversion. Image height $h_i = m \times h_o$:

The magnitude $5.0\ \text{cm}$ tells the size; the negative sign confirms inversion.

Step 6: Describe the image.

• Real (can be projected on a screen)
• Inverted (upside down)
• Magnified ($5.0\ \text{cm}$ tall)
• Formed 54 cm in front of the mirror

Step 7: What happens when the candle is moved closer?

If the object moves toward the mirror (i.e., $|u|$ decreases), the image distance $v$ changes. Let's examine two cases:

• Object beyond $C$ ($|u| > 36\ \text{cm}$): Image between $C$ and $F$, real and smaller.
• Object between $C$ and $F$ ($18 < |u| < 36$): Image beyond $C$, real and enlarged.
• Object at $F$ ($|u| = 18\ \text{cm}$): Image at infinity — no sharp image on any screen.
• Object between $F$ and pole ($|u| < 18\ \text{cm}$): Image virtual, behind the mirror — cannot be caught on a screen.

In our problem, the candle is at $27\ \text{cm}$ (between $C$ and $F$). Moving it closer to the mirror means $|u|$ decreases from $27$ toward $18\ \text{cm}$. From the mirror formula:

As $|u|$ decreases, $\frac{1}{|u|}$ increases, so $\frac{1}{v}$ becomes more negative — meaning $|v|$ increases. The screen must be moved farther away from the mirror.

When the candle reaches $18\ \text{cm}$ (the focal point), $v \to \infty$ — no image on any screen. Beyond that, the image becomes virtual and the screen is useless.

Method: Mirror Formula and Magnification Approach

This problem uses the Mirror Formula combined with Magnification relations — the standard method for all spherical mirror image-location problems in ray optics.

Step 1: Identify given data and sign convention

Sign convention: Cartesian sign convention (all distances measured from pole, direction of incident light is positive).

• Object size, $h_o = +2.5\ \text{cm}$
• Object distance, $u = -27\ \text{cm}$ (negative: object is in front of mirror)
• Radius of curvature, $R = -36\ \text{cm}$ (negative for concave mirror)

Step 2: Find focal length

For a spherical mirror:

Step 3: Apply mirror formula to find image distance

Mirror formula:

Substitute $u = -27\ \text{cm}$, $f = -18\ \text{cm}$:

Take LCM (54):

Screen should be placed 54 cm in front of the mirror.

Step 4: Find magnification and image size

Magnification:

Image size:

Step 5: Describe the nature of the image

• Real (since $v$ is negative, image is in front of mirror)
• Inverted (since $m$ is negative)
• Magnified (since $|m| > 1$)
• Size: 5 cm (inverted)

Step 6: Effect of moving candle closer

If the candle is moved closer to the mirror (i.e., $|u|$ decreases):

• For a concave mirror, as object moves from beyond $C$ toward $F$, the image moves farther away from the mirror (beyond $C$).
• The screen must be moved away from the mirror to capture the sharp image.

Rule: For a concave mirror, object moving toward mirror → image moves away from mirror (until object reaches focus, after which image becomes virtual).

Final Answer Summary

Here are the common mistakes students make with this exact problem, and how to avoid each one.

1. Confusing Radius of Curvature ($R$) with Focal Length ($f$)

The Mistake:

Students plug $R = 36\ \text{cm}$ directly into the mirror formula as $f$.

Why it’s wrong:

For a spherical mirror, the focal length is half the radius of curvature:

How to avoid:

Always write the relation first:

Then use $f = -18\ \text{cm}$ (concave mirror → negative focal length as per Cartesian sign convention).

2. Forgetting the Sign Convention (Cartesian)

The Mistake:

Using $u = 27\ \text{cm}$ as positive, or treating $f$ as positive.

Why it’s wrong:

In the Cartesian sign convention:

• Distances measured against the incident light direction are negative.
• For a concave mirror, both $u$ and $f$ are negative.

How to avoid:

Always draw a quick ray diagram and label directions. Then:

3. Using the Lens Formula Instead of the Mirror Formula

The Mistake:

Writing $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ (lens formula) instead of the mirror formula.

Why it’s wrong:

The mirror formula is:

How to avoid:

Memorise separately:

• Mirror: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
• Lens: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

Write the correct one before substituting.

4. Sign Error When Solving for $v$

The Mistake:

After substituting $u = -27$ and $f = -18$, students get the algebra wrong and end up with $v$ positive when it should be negative (or vice versa).

How to avoid:

Do the algebra step-by-step:

Find LCM = 54:

Thus:

Check: Negative $v$ means the image is in front of the mirror (real image), which matches the problem’s requirement for a screen.

5. Misinterpreting Magnification Sign

The Mistake:

Using $m = \frac{v}{u}$ without signs, or forgetting that $m$ negative means inverted image.

How to avoid:

Use the signed formula:

• Negative $m$ → image is inverted.
• $|m| = 2$ → image is magnified (twice the object size).

6. Calculating Image Size Incorrectly

The Mistake:

Using $h_i = m \times h_o$ but forgetting the sign of $m$ or $h_o$.

How to avoid:

Object size $h_o = +2.5\ \text{cm}$ (always positive). Then:

• $|h_i| = 5.0\ \text{cm}$ → size.
• Negative sign → image is inverted relative to object.

7. Describing the Image Incorrectly

The Mistake:

Saying “real, inverted, diminished” when it’s actually magnified.

How to avoid:

Check $|m|$:

• $|m| > 1$ → magnified
• $|m| < 1$ → diminished
• $|m| = 1$ → same size

Here $|m| = 2$, so the image is real, inverted, and magnified.

8. Answering the “Move the Screen” Part Wrong

The Mistake:

Saying “move the screen away” when the candle is moved closer, without checking the new $u$.

How to avoid:

If the candle is moved closer (say $u$ becomes less negative, e.g., $-20\ \text{cm}$), recalculate $v$:

So $v = -180\ \text{cm}$ — farther from the mirror.

Key rule: For a concave mirror, as the object moves from infinity toward the focus, the image moves away from the mirror (from focus to infinity). So the screen must be moved farther from the mirror.

Quick Summary Checklist

Final Answer for the problem:

• Screen distance: $54\ \text{cm}$ in front of the mirror
• Image: real, inverted, $5.0\ \text{cm}$ tall
• If candle is moved closer: move screen farther from the mirror