How to Solve: Lens Formula

How to Solve: Lens Formula

(For Students Who Want to Ace Their Exam & Teachers Who Need a Ready-to-Record Script)

Introduction

"Imagine you’re designing glasses for a patient, or a telescope to see distant stars—without the lens formula, you’re just guessing. Master this, and you’ll solve any lens problem in your exam, guaranteed."

What You Need To Know First

Before diving into the lens formula, ensure you understand: 1. Sign conventions – How to assign positive/negative values to object distance (u), image distance (v), and focal length (f). 2. Ray diagrams – How light bends through converging (convex) and diverging (concave) lenses. 3. Real vs. virtual images – Real images form on the opposite side of the lens (positive v); virtual images form on the same side (negative v).

Key Vocabulary

Formulas To Know

1. Lens Formula

Formula: [ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} ] Variables: - f = Focal length of the lens (cm or m). - v = Image distance from the lens (cm or m). - u = Object distance from the lens (cm or m).

Sign Rules (CRUCIAL!): - Convex lens (converging): f = positive. - Concave lens (diverging): f = negative. - Object distance (u): Always negative (object is on the left side of the lens). - Image distance (v): Positive if real image (opposite side of lens); negative if virtual image (same side as object).

MEMORISE THIS? ✅ Yes! (Not always given on exam sheets.)

2. Magnification Formula

Formula: [ m = \frac{v}{u} = \frac{\text{Height of image (h₂)}}{\text{Height of object (h₁)}} ] Variables: - m = Magnification (no units). - v = Image distance. - u = Object distance.

Sign Rules: - m = Positive → Image is upright (virtual). - m = Negative → Image is inverted (real).

MEMORISE THIS? ✅ Yes!

Step-by-Step Method

Follow these steps exactly for every lens problem.

Step 1: Identify the lens type

• Convex lens? → f = positive.
• Concave lens? → f = negative.

Step 2: Assign signs to given values

• Object distance (u): Always negative (object is on the left).
• Image distance (v): Positive if real (opposite side); negative if virtual (same side).
• Focal length (f): As above (positive for convex, negative for concave).

Step 3: Plug values into the lens formula

[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} ] Rearrange to solve for the unknown (v or u or f).

Step 4: Solve for the unknown

• Use algebra to isolate the unknown.
• Pro tip: If solving for v or u, take reciprocals at the end.

Step 5: Check the sign of your answer

• Positive v? → Real image (opposite side of lens).
• Negative v? → Virtual image (same side as object).
• Magnification (m): Negative? → Inverted image.

Step 6: Verify with magnification (if needed)

[ m = \frac{v}{u} ] - Compare with image height if given.

Worked Examples

Example 1 – Basic (Convex Lens)

Problem: An object is placed 30 cm in front of a convex lens with a focal length of 10 cm. Find the image distance (v) and magnification (m).

Step-by-Step Solution:

1. Identify lens type: Convex → f = +10 cm.
2. Assign signs:
3. u = -30 cm (object on left).
4. f = +10 cm.
5. Plug into lens formula:
   [
   \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \
   \frac{1}{10} = \frac{1}{v} - \frac{1}{-30} \
   \frac{1}{10} = \frac{1}{v} + \frac{1}{30}
   ]
6. Solve for v:
   [
   \frac{1}{v} = \frac{1}{10} - \frac{1}{30} = \frac{3}{30} - \frac{1}{30} = \frac{2}{30} = \frac{1}{15} \
   v = +15 \text{ cm}
   ]
7. Interpret v:
8. v = +15 cm → Real image, 15 cm behind the lens.
9. Find magnification (m):
   [
   m = \frac{v}{u} = \frac{15}{-30} = -0.5
   ]
10. m = -0.5 → Image is inverted and half the size of the object.

What we did and why: - Used the lens formula to find v. - Applied sign rules correctly (negative u, positive f). - Calculated magnification to describe the image.

Example 2 – Medium (Concave Lens)

Problem: An object is placed 20 cm in front of a concave lens with a focal length of 15 cm. Find the image distance (v) and describe the image.

Step-by-Step Solution:

1. Identify lens type: Concave → f = -15 cm.
2. Assign signs:
3. u = -20 cm.
4. f = -15 cm.
5. Plug into lens formula:
   [
   \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \
   \frac{1}{-15} = \frac{1}{v} - \frac{1}{-20} \
   -\frac{1}{15} = \frac{1}{v} + \frac{1}{20}
   ]
6. Solve for v:
   [
   \frac{1}{v} = -\frac{1}{15} - \frac{1}{20} = -\frac{4}{60} - \frac{3}{60} = -\frac{7}{60} \
   v = -\frac{60}{7} \approx -8.57 \text{ cm}
   ]
7. Interpret v:
8. v = -8.57 cm → Virtual image, 8.57 cm in front of the lens (same side as object).
9. Find magnification (m):
   [
   m = \frac{v}{u} = \frac{-8.57}{-20} = +0.4285
   ]
10. m = +0.43 → Image is upright and smaller than the object.

What we did and why: - Used negative f for a concave lens. - Solved for v and got a negative value → virtual image. - Magnification confirmed the image is upright and reduced.

Example 3 – Exam Style (Disguised Problem)

Problem: A 5 cm tall object is placed 12 cm from a lens. The image formed is virtual, upright, and 2.5 cm tall. Find the focal length of the lens.

Step-by-Step Solution:

1. Determine lens type:
2. Virtual, upright image → Concave lens (convex lenses form real, inverted images unless object is inside f).
3. So, f = negative.
4. Find magnification (m):
   [
   m = \frac{\text{Height of image}}{\text{Height of object}} = \frac{2.5}{5} = +0.5
   ]
5. m = +0.5 → Upright image.
6. Relate m to v and u:
   [
   m = \frac{v}{u} \
   0.5 = \frac{v}{-12} \quad (\text{u is negative}) \
   v = -6 \text{ cm}
   ]
7. v = -6 cm → Virtual image (same side as object).
8. Plug into lens formula:
   [
   \frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{-6} - \frac{1}{-12} = -\frac{1}{6} + \frac{1}{12} = -\frac{2}{12} + \frac{1}{12} = -\frac{1}{12} \
   f = -12 \text{ cm}
   ]
9. Interpret f:
10. f = -12 cm → Concave lens with focal length 12 cm.

What we did and why: - Used image properties to deduce lens type. - Calculated m from image/object heights. - Solved for f using the lens formula.

Common Mistakes

Exam Traps

1-Minute Recap

"Alright, let’s lock this in—30 seconds to exam success!

1. Lens formula: ( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} ). Memorise it.
2. Signs are everything:
3. u = always negative (object on left).
4. f = positive for convex, negative for concave.
5. v = positive for real images, negative for virtual.
6. Magnification: ( m = \frac{v}{u} ). Negative m? Image is inverted.
7. If stuck: Draw a quick ray diagram to visualise the image.
8. Double-check: Did you assign signs correctly? Did you take reciprocals properly?

You’ve got this. Now go solve that lens problem like a pro!