Class 10 Science Mirror formula and Refractive Index Numerical Worksheet Chapter 9 Light Reflection and Refraction | Detailed Solutions & Marking Scheme
๐ VEDANT IGNITE WORKSHEET
Class 10 Physics – Light (Numericals Practice)
Level: Moderate to Advanced | Marks: 30
๐ท Important Formulae (Revise Before Solving)
Mirror Formula
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Magnification (Mirror)
m = \frac{h_i}{h_o} = -\frac{v}{u}
Refractive Index
n = \frac{c}{v}
๐ง Section A – Basic to Moderate (1–2 Marks Each)
Q1.
An object is placed at a distance of –30 cm from a concave mirror of focal length –15 cm.
Find the position of the image.
Q2.
A convex mirror has focal length +20 cm. An object is placed at –40 cm.
Find image distance and nature of image.
Q3.
An image formed by a mirror is twice the size of the object and erect.
Find magnification and identify the type of mirror.
Q4.
Speed of light in a medium is 2 × 10⁸ m/s.
Calculate its refractive index.
Q5.
An object of height 4 cm forms an image of height –8 cm.
Find magnification and nature of image.
๐ท Section B – Competency-Based (3 Marks Each)
Q6. (Application-Based)
A dentist uses a mirror to see enlarged images of teeth.
If the image is formed at –10 cm and object is at –5 cm:
Find focal length
Identify mirror type
Justify why this mirror is used
Q7. (Real-Life Scenario)
A car uses a convex mirror as rear-view mirror.
An object is placed at –100 cm, focal length = +50 cm
Find image position
Calculate magnification
Explain why convex mirrors are preferred
Q8. (Graphical Thinking)
An object is moved from infinity towards a concave mirror.
At what position will image be formed when object is at 2f?
Find magnification at this point numerically
Q9. (Concept Integration)
Light travels from air into a medium where its speed becomes 1.5 × 10⁸ m/s
Find refractive index
Compare optical density with air
State effect on speed and wavelength
๐ด Section C – Higher Order Thinking (4 Marks Each)
Q10. (Case Study)
A student performs an experiment with a concave mirror:
| Object Distance (cm) | Image Distance (cm) |
|---|---|
| –20 | –20 |
| –30 | –15 |
Find focal length using both cases
Comment on consistency
Identify error (if any)
Q11. (Multi-Step Numerical)
An object is placed 15 cm in front of a concave mirror of focal length 10 cm
Find image distance
Calculate magnification
Find image height if object height = 2 cm
State nature of image
Q12. (Mixed Concept Problem)
A ray of light enters a medium with refractive index 1.5
Find speed of light in that medium
If object is placed in front of a mirror in this medium, does mirror formula change? Justify
Explain difference between optical density and mass density
⚡ Section D – Challenge Questions (5 Marks Each)
Q13. (Advanced Reasoning)
An object is placed between F and C of a concave mirror.
The image formed is 3 times enlarged.
Find object distance in terms of focal length
Find image distance
Verify using magnification formula
Q14. (Concept Twist)
Two mirrors are used:
Mirror A: Concave (f = –20 cm)
Mirror B: Convex (f = +20 cm)
An object is placed at –30 cm from both mirrors
Compare image positions
Compare magnifications
Which mirror gives wider field of view and why?
Q15. (Experimental Thinking)
A student gets wrong image distance repeatedly.
Possible reasons:
Incorrect sign convention
Measuring from pole incorrectly
Mirror not aligned properly
Using a numerical example, show how sign error affects final answer.
Here’s a detailed Answer Key with stepwise solutions + marking scheme for your worksheet ๐
๐ VEDANT IGNITE – ANSWER KEY
Class 10 Physics – Light (Numericals)
๐ท Section A (1–2 Marks Each)
Q1.
Given:
u = –30 cm, f = –15 cm
Using mirror formula:
1/f = 1/v + 1/u
1/–15 = 1/v + 1/–30
⇒ 1/v = –1/15 + 1/30 = –1/30
⇒ v = –30 cm
Answer: Image at –30 cm (real, inverted, same size)
Marks: 2 (formula + substitution + answer)
Q2.
u = –40 cm, f = +20 cm
1/20 = 1/v – 1/40
⇒ 1/v = 1/20 + 1/40 = 3/40
⇒ v = +13.3 cm
Answer: Image at +13.3 cm, virtual, erect, diminished
Marks: 2
Q3.
Image is erect and twice size
⇒ m = +2
Only convex mirror cannot give magnification >1, so it must be concave mirror (object between pole and focus)
Answer: m = +2, concave mirror
Marks: 2
Q4.
n = c/v = (3 × 10⁸)/(2 × 10⁸) = 1.5
Answer: 1.5
Marks: 1
Q5.
m = hแตข/h₀ = –8/4 = –2
Answer: m = –2 → real, inverted, enlarged
Marks: 2
๐ท Section B (3 Marks Each)
Q6.
u = –5 cm, v = –10 cm
1/f = 1/v + 1/u
= –1/10 – 1/5 = –3/10
⇒ f = –10/3 ≈ –3.33 cm
Mirror: Concave
Justification: Produces enlarged, erect image when object is close
Marks:
Formula + calc: 1
Mirror type: 1
Justification: 1
Q7.
u = –100 cm, f = +50 cm
1/50 = 1/v – 1/100
⇒ 1/v = 3/100
⇒ v = +33.3 cm
m = –v/u = –(33.3/–100) = +0.33
Answer:
Image at +33.3 cm
m = +0.33 (diminished, erect)
Reason: wider field of view
Marks: 3
Q8.
At object at 2f → image also at 2f
m = –v/u = –(2f/–2f) = –1
Answer: Image at 2f, m = –1 (same size, inverted)
Marks: 3
Q9.
n = (3 × 10⁸)/(1.5 × 10⁸) = 2
More optically dense than air
Speed decreases
Wavelength decreases
Marks:
Calculation: 1
Density: 1
Effect: 1
๐ด Section C (4 Marks Each)
Q10.
Case 1:
u = –20, v = –20
⇒ f = –10 cm
Case 2:
u = –30, v = –15
⇒ f = –10 cm
Consistency: Yes
Error: No major error (values correct)
Marks:
Case 1: 1
Case 2: 1
Conclusion: 1
Error analysis: 1
Q11.
u = –15 cm, f = –10 cm
1/–10 = 1/v – 1/15
⇒ 1/v = –1/10 + 1/15 = –1/30
⇒ v = –30 cm
m = –v/u = –(–30/–15) = –2
hแตข = m × h₀ = –2 × 2 = –4 cm
Answer:
v = –30 cm
m = –2
hแตข = –4 cm
Real, inverted, enlarged
Marks: 4
Q12.
Speed = c/n = (3 × 10⁸)/1.5 = 2 × 10⁸ m/s
Mirror formula → No change (independent of medium)
Difference:
Optical density → depends on refractive index
Mass density → depends on mass/volume
Marks: 4
⚡ Section D (5 Marks Each)
Q13.
m = –v/u = 3
⇒ v = –3u
Using mirror formula:
1/f = 1/v + 1/u
⇒ 1/f = 1/(–3u) + 1/u
= (–1 + 3)/3u = 2/3u
⇒ u = 2f/3
v = –3u = –2f
Answer:
u = 2f/3
v = –2f
Verified
Marks: 5
Q14.
Concave Mirror:
v ≈ –60 cm (real, inverted)
m ≈ –2
Convex Mirror:
v ≈ +12 cm
m ≈ +0.4
Field of View: Convex mirror (wider)
Marks:
Concave: 2
Convex: 2
Explanation: 1
Q15.
Example:
Correct:
u = –20, f = –10 → v = –20
Wrong sign:
u = +20
⇒ wrong v → incorrect nature
Conclusion: Sign convention changes result drastically
Marks:
Example: 2
Wrong calc: 2
Conclusion: 1
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