Class 10 Physics – Tough Numerical Practice
Sign Convention (Plane & Spherical Mirrors)
- A concave mirror forms a real image three times the size of the object at a distance of 36 cm from the mirror.
Find:
(a) Image distance
(b) Object distance
(c) Focal length
(d) Nature of image
Hint: Real image → image distance negative. Inverted image → magnification negative. - An object is placed at 15 cm from a concave mirror and produces a virtual image at 30 cm.
Find focal length and magnification.
Hint: Virtual image → image distance positive. Concave mirror → focal length negative. - A convex mirror forms an image half the size of the object at 20 cm behind the mirror.
Find object distance and focal length.
Hint: Convex mirror → focal length positive. Image behind mirror → positive image distance. - An object placed in front of a concave mirror produces an image at the same distance as the object.
Find the position of object in terms of focal length.
Hint: Use mirror formula and consider real image case. - A concave mirror of focal length 25 cm produces an image at 50 cm.
Find object distance and magnification.
Hint: Decide sign based on image type before substituting. - A convex mirror has radius of curvature 40 cm.
An object is placed 30 cm in front of it.
Find image distance and magnification.
Hint: f = R/2. Convex mirror → positive focal length. - A concave mirror produces a magnification of –4. If image is formed at 40 cm in front of mirror, find object distance and focal length.
- An object of height 5 cm is placed 20 cm in front of a convex mirror of focal length 15 cm. Find image height and nature of image.
- A concave mirror of radius 60 cm forms an image 20 cm from mirror. Find object distance and magnification.
- An object is placed between pole and focus of a concave mirror of focal length 12 cm. If object distance is 8 cm, find image distance and magnification.
Class 10 Physics – Sign Convention Numerical Practice
Plane & Spherical Mirrors (CBSE Level – With Sign Hints)
Set 1
- An object is placed 24 cm in front of a concave mirror of focal length 12 cm. Find image distance and magnification.
Sign Hint: Object in front → \(u\) negative. Concave mirror → \(f\) negative. Real image (if formed in front) → \(v\) negative. - A convex mirror of focal length 20 cm forms an image when object is at 30 cm. Find image distance.
Sign Hint: Convex mirror → \(f\) positive. Object in front → \(u\) negative. Convex mirror forms virtual image → \(v\) positive. - An object is placed at 10 cm from a plane mirror. Find image position.
Sign Hint: Object in front → \(u\) negative. Image behind mirror → \(v\) positive. - A concave mirror forms a real image at 36 cm when object is placed at 18 cm. Find focal length.
Sign Hint: Real image in front → \(v\) negative. Object in front → \(u\) negative. Concave mirror → \(f\) negative. - A convex mirror forms image 15 cm behind mirror. If focal length is 25 cm, find object distance.
Sign Hint: Image behind mirror → \(v\) positive. Convex mirror → \(f\) positive.
Set 2
- An object is placed at 30 cm in front of concave mirror of focal length 15 cm. Find image position.
Sign Hint: Object in front → negative. Concave mirror → \(f\) negative. - A convex mirror of radius 40 cm forms image at 10 cm. Find object distance.
Sign Hint: Radius of convex mirror → positive. \(f = R/2\). Image behind mirror → positive. - A concave mirror produces magnification –2. If image distance is 20 cm, find object distance.
Sign Hint: Negative magnification → inverted → real image → \(v\) negative. - An object is placed 12 cm from plane mirror. Where is image formed?
Sign Hint: Plane mirror image distance equals object distance but positive. - A concave mirror forms virtual image at 25 cm. Find object distance if focal length is 20 cm.
Sign Hint: Virtual image → behind mirror → \(v\) positive. Concave mirror → \(f\) negative.
Set 3
- A concave mirror of focal length 18 cm forms image at 9 cm. Find object distance.
Sign Hint: Concave mirror → \(f\) negative. Decide sign of \(v\) based on image nature. - A convex mirror forms image half the size of object at 16 cm. Find focal length.
Sign Hint: Convex mirror → \(f\) positive. Image always virtual → \(v\) positive. - An object placed at focus of concave mirror. Where is image formed?
Sign Hint: Object at focus → \(u = f\) (both negative). - A concave mirror produces image equal in size to object. Find object position.
Sign Hint: Equal size real image → object at centre of curvature. - Object placed at 20 cm in front of convex mirror of focal length 15 cm. Find image position.
Sign Hint: Object negative, focal length positive.
Set 4
- A concave mirror produces real image 3 times magnified at 30 cm. Find object distance.
Sign Hint: Real image → inverted → magnification negative. - A convex mirror forms image 12 cm behind mirror when object is at 24 cm. Find focal length.
Sign Hint: Both distances follow sign rule before substitution. - Plane mirror produces image 8 cm behind mirror. Find object position.
Sign Hint: Image positive → object equal magnitude but negative. - Concave mirror radius is 50 cm. Find focal length and its sign.
Sign Hint: Concave mirror → focal length negative. - A concave mirror produces virtual image 15 cm behind mirror. Find magnification.
Sign Hint: Virtual image → positive \(v\). Magnification positive for erect image.
Set 5
- A concave mirror of focal length 10 cm forms image at 20 cm. Find object distance.
Sign Hint: Check if image is real or virtual before assigning sign. - A convex mirror of focal length 30 cm forms image at 10 cm. Find object distance.
Sign Hint: Convex → \(f\) positive. Image behind mirror → positive. - An object 15 cm from plane mirror. Find image distance and magnification.
Sign Hint: Plane mirror → \(m = +1\). - A concave mirror forms image at 40 cm for object at 20 cm. Find focal length.
Sign Hint: Both object and real image in front → negative values. - A convex mirror produces image of height 2 cm for object 8 cm placed at 32 cm. Find focal length.
Sign Hint: Magnification positive for convex mirror.
Solutions – Tough Numerical Practice (Mirrors)
1.
Real image three times the size at 36 cm
Real image → \( v = -36\,cm \)
Magnification \( m = -3 \)
\[
m = -\frac{v}{u}
\]
\[
-3 = -\frac{-36}{u}
\]
\[
-3 = \frac{36}{u}
\]
\[
u = -12\,cm
\]
Mirror formula:
\[
\frac{1}{f} = \frac{1}{v} + \frac{1}{u}
\]
\[
\frac{1}{f} = \frac{1}{-36} + \frac{1}{-12}
\]
\[
\frac{1}{f} = \frac{-1-3}{36}
\]
\[
f = -9\,cm
\]
Image is real and inverted.
2. Virtual image → \( v = +30\,cm \) Object → \( u = -15\,cm \) \[ \frac{1}{f} = \frac{1}{30} + \frac{1}{-15} \] \[ \frac{1}{f} = \frac{1-2}{30} \] \[ f = -30\,cm \] \[ m = -\frac{30}{-15} = +2 \]
3. Convex mirror \( v = +20\,cm \) \( m = 0.5 \) \[ 0.5 = -\frac{20}{u} \] \[ u = -40\,cm \] \[ \frac{1}{f} = \frac{1}{20} + \frac{1}{-40} \] \[ f = +40\,cm \]
4. Image at same distance as object For real image: \[ v = u \] From mirror formula: \[ \frac{1}{f} = \frac{2}{u} \] \[ u = 2f \] Object is at centre of curvature.
5. \( f = -25\,cm \) Image at 50 cm (real) → \( v = -50\,cm \) \[ \frac{1}{-25} = \frac{1}{-50} + \frac{1}{u} \] \[ \frac{1}{u} = \frac{-2+1}{50} \] \[ u = -50\,cm \] \[ m = -\frac{-50}{-50} = -1 \]
6. Convex mirror \( R = 40\,cm \) \[ f = \frac{R}{2} = +20\,cm \] \( u = -30\,cm \) \[ \frac{1}{20} = \frac{1}{v} + \frac{1}{-30} \] \[ \frac{1}{v} = \frac{1}{20} + \frac{1}{30} \] \[ \frac{1}{v} = \frac{3+2}{60} \] \[ v = +12\,cm \] \[ m = -\frac{12}{-30} = +0.4 \]
7. \( m = -4 \) \( v = -40\,cm \) \[ -4 = -\frac{-40}{u} \] \[ u = -10\,cm \] \[ \frac{1}{f} = \frac{1}{-40} + \frac{1}{-10} \] \[ \frac{1}{f} = \frac{-1-4}{40} \] \[ f = -8\,cm \]
8. Convex mirror \( u = -20\,cm \) \( f = +15\,cm \) \[ \frac{1}{15} = \frac{1}{v} + \frac{1}{-20} \] \[ \frac{1}{v} = \frac{1}{15} + \frac{1}{20} \] \[ v = +8.6\,cm \] \[ m = -\frac{8.6}{-20} = +0.43 \] \[ h_i = 0.43 \times 5 = 2.15\,cm \] Virtual and erect.
9. Concave mirror \( R = -60\,cm \) \[ f = -30\,cm \] \( v = -20\,cm \) \[ \frac{1}{-30} = \frac{1}{-20} + \frac{1}{u} \] \[ \frac{1}{u} = \frac{-2+3}{60} \] \[ u = +60\,cm \] \[ m = -\frac{-20}{60} = 0.33 \]
10. Object inside focus \( u = -8\,cm \) \( f = -12\,cm \) \[ \frac{1}{-12} = \frac{1}{v} + \frac{1}{-8} \] \[ \frac{1}{v} = \frac{-1+1.5}{12} \] \[ v = +24\,cm \] \[ m = -\frac{24}{-8} = +3 \] Virtual and magnified.
2. Virtual image → \( v = +30\,cm \) Object → \( u = -15\,cm \) \[ \frac{1}{f} = \frac{1}{30} + \frac{1}{-15} \] \[ \frac{1}{f} = \frac{1-2}{30} \] \[ f = -30\,cm \] \[ m = -\frac{30}{-15} = +2 \]
3. Convex mirror \( v = +20\,cm \) \( m = 0.5 \) \[ 0.5 = -\frac{20}{u} \] \[ u = -40\,cm \] \[ \frac{1}{f} = \frac{1}{20} + \frac{1}{-40} \] \[ f = +40\,cm \]
4. Image at same distance as object For real image: \[ v = u \] From mirror formula: \[ \frac{1}{f} = \frac{2}{u} \] \[ u = 2f \] Object is at centre of curvature.
5. \( f = -25\,cm \) Image at 50 cm (real) → \( v = -50\,cm \) \[ \frac{1}{-25} = \frac{1}{-50} + \frac{1}{u} \] \[ \frac{1}{u} = \frac{-2+1}{50} \] \[ u = -50\,cm \] \[ m = -\frac{-50}{-50} = -1 \]
6. Convex mirror \( R = 40\,cm \) \[ f = \frac{R}{2} = +20\,cm \] \( u = -30\,cm \) \[ \frac{1}{20} = \frac{1}{v} + \frac{1}{-30} \] \[ \frac{1}{v} = \frac{1}{20} + \frac{1}{30} \] \[ \frac{1}{v} = \frac{3+2}{60} \] \[ v = +12\,cm \] \[ m = -\frac{12}{-30} = +0.4 \]
7. \( m = -4 \) \( v = -40\,cm \) \[ -4 = -\frac{-40}{u} \] \[ u = -10\,cm \] \[ \frac{1}{f} = \frac{1}{-40} + \frac{1}{-10} \] \[ \frac{1}{f} = \frac{-1-4}{40} \] \[ f = -8\,cm \]
8. Convex mirror \( u = -20\,cm \) \( f = +15\,cm \) \[ \frac{1}{15} = \frac{1}{v} + \frac{1}{-20} \] \[ \frac{1}{v} = \frac{1}{15} + \frac{1}{20} \] \[ v = +8.6\,cm \] \[ m = -\frac{8.6}{-20} = +0.43 \] \[ h_i = 0.43 \times 5 = 2.15\,cm \] Virtual and erect.
9. Concave mirror \( R = -60\,cm \) \[ f = -30\,cm \] \( v = -20\,cm \) \[ \frac{1}{-30} = \frac{1}{-20} + \frac{1}{u} \] \[ \frac{1}{u} = \frac{-2+3}{60} \] \[ u = +60\,cm \] \[ m = -\frac{-20}{60} = 0.33 \]
10. Object inside focus \( u = -8\,cm \) \( f = -12\,cm \) \[ \frac{1}{-12} = \frac{1}{v} + \frac{1}{-8} \] \[ \frac{1}{v} = \frac{-1+1.5}{12} \] \[ v = +24\,cm \] \[ m = -\frac{24}{-8} = +3 \] Virtual and magnified.
Solutions – Sign Convention (Plane & Spherical Mirrors)
1.
Given: \( u = -24\,cm \), \( f = -12\,cm \)
\[
\frac{1}{f} = \frac{1}{v} + \frac{1}{u}
\]
\[
\frac{1}{-12} = \frac{1}{v} + \frac{1}{-24}
\]
\[
\frac{1}{v} = \frac{1}{-12} – \frac{1}{-24}
\]
\[
\frac{1}{v} = \frac{-2+1}{24} = \frac{-1}{24}
\]
\[
v = -24\,cm
\]
\[
m = -\frac{v}{u} = -\frac{-24}{-24} = -1
\]
2. Given: \( u = -30\,cm \), \( f = +20\,cm \) \[ \frac{1}{20} = \frac{1}{v} + \frac{1}{-30} \] \[ \frac{1}{v} = \frac{1}{20} + \frac{1}{30} \] \[ \frac{1}{v} = \frac{3+2}{60} = \frac{5}{60} \] \[ v = 12\,cm \]
3. Plane mirror Image distance = +10 cm \[ m = +1 \]
4. Given: \( u = -18\,cm \), \( v = -36\,cm \) \[ \frac{1}{f} = \frac{1}{-36} + \frac{1}{-18} \] \[ \frac{1}{f} = \frac{-1-2}{36} \] \[ f = -12\,cm \]
5. Given: \( v = +15\,cm \), \( f = +25\,cm \) \[ \frac{1}{25} = \frac{1}{15} + \frac{1}{u} \] \[ \frac{1}{u} = \frac{1}{25} – \frac{1}{15} \] \[ \frac{1}{u} = \frac{3-5}{75} = \frac{-2}{75} \] \[ u = -37.5\,cm \]
6. Given: \( u = -30\,cm \), \( f = -15\,cm \) \[ \frac{1}{-15} = \frac{1}{v} + \frac{1}{-30} \] \[ \frac{1}{v} = \frac{-2+1}{30} \] \[ v = -30\,cm \]
7. \( R = +40\,cm \) \[ f = \frac{R}{2} = +20\,cm \]
8. Given: \( m = -2 \), \( v = -20\,cm \) \[ -2 = -\frac{v}{u} \] \[ -2 = -\frac{-20}{u} \] \[ u = -10\,cm \]
9. Plane mirror Image distance = +12 cm \[ m = +1 \]
10. Given: \( v = +25\,cm \), \( f = -20\,cm \) \[ \frac{1}{-20} = \frac{1}{25} + \frac{1}{u} \] \[ \frac{1}{u} = \frac{-5-4}{100} \] \[ u = -11.1\,cm \]
11. Given: \( f = -18\,cm \), \( v = -9\,cm \) \[ \frac{1}{-18} = \frac{1}{-9} + \frac{1}{u} \] \[ \frac{1}{u} = \frac{-1+2}{18} \] \[ u = +18\,cm \]
12. Given: \( m = 0.5 \), \( v = +16\,cm \) \[ 0.5 = -\frac{16}{u} \] \[ u = -32\,cm \] \[ \frac{1}{f} = \frac{1}{16} + \frac{1}{-32} \] \[ f = +32\,cm \]
13. Object at focus \[ v = \infty \]
14. Equal size real image \[ u = 2f \]
15. Given: \( u = -20\,cm \), \( f = +15\,cm \) \[ \frac{1}{15} = \frac{1}{v} + \frac{1}{-20} \] \[ v = 60\,cm \]
16. Given: \( m = -3 \), \( v = -30\,cm \) \[ -3 = -\frac{-30}{u} \] \[ u = -10\,cm \]
17. Given: \( u=-24\,cm \), \( v=+12\,cm \) \[ \frac{1}{f} = \frac{1}{12} + \frac{1}{-24} \] \[ f = +24\,cm \]
18. Plane mirror Object = -8 cm Image = +8 cm
19. \( f = -25\,cm \)
20. Given: \( v = +15\,cm \) \[ m = -\frac{15}{u} \] \[ m > 0 \]
21. Given: \( f=-10\,cm \), \( v=-20\,cm \) \[ \frac{1}{-10} = \frac{1}{-20} + \frac{1}{u} \] \[ u = -20\,cm \]
22. Given: \( f=+30\,cm \), \( v=+10\,cm \) \[ \frac{1}{30} = \frac{1}{10} + \frac{1}{u} \] \[ u = -15\,cm \]
23. Plane mirror Image = +15 cm \[ m = +1 \]
24. Given: \( u=-20\,cm \), \( v=-40\,cm \) \[ \frac{1}{f} = \frac{1}{-40} + \frac{1}{-20} \] \[ f = -13.3\,cm \]
25. Given: \( m = \frac{2}{8} = 0.25 \) \[ 0.25 = -\frac{v}{u} \] Solve to get: \[ f = +32\,cm \]
2. Given: \( u = -30\,cm \), \( f = +20\,cm \) \[ \frac{1}{20} = \frac{1}{v} + \frac{1}{-30} \] \[ \frac{1}{v} = \frac{1}{20} + \frac{1}{30} \] \[ \frac{1}{v} = \frac{3+2}{60} = \frac{5}{60} \] \[ v = 12\,cm \]
3. Plane mirror Image distance = +10 cm \[ m = +1 \]
4. Given: \( u = -18\,cm \), \( v = -36\,cm \) \[ \frac{1}{f} = \frac{1}{-36} + \frac{1}{-18} \] \[ \frac{1}{f} = \frac{-1-2}{36} \] \[ f = -12\,cm \]
5. Given: \( v = +15\,cm \), \( f = +25\,cm \) \[ \frac{1}{25} = \frac{1}{15} + \frac{1}{u} \] \[ \frac{1}{u} = \frac{1}{25} – \frac{1}{15} \] \[ \frac{1}{u} = \frac{3-5}{75} = \frac{-2}{75} \] \[ u = -37.5\,cm \]
6. Given: \( u = -30\,cm \), \( f = -15\,cm \) \[ \frac{1}{-15} = \frac{1}{v} + \frac{1}{-30} \] \[ \frac{1}{v} = \frac{-2+1}{30} \] \[ v = -30\,cm \]
7. \( R = +40\,cm \) \[ f = \frac{R}{2} = +20\,cm \]
8. Given: \( m = -2 \), \( v = -20\,cm \) \[ -2 = -\frac{v}{u} \] \[ -2 = -\frac{-20}{u} \] \[ u = -10\,cm \]
9. Plane mirror Image distance = +12 cm \[ m = +1 \]
10. Given: \( v = +25\,cm \), \( f = -20\,cm \) \[ \frac{1}{-20} = \frac{1}{25} + \frac{1}{u} \] \[ \frac{1}{u} = \frac{-5-4}{100} \] \[ u = -11.1\,cm \]
11. Given: \( f = -18\,cm \), \( v = -9\,cm \) \[ \frac{1}{-18} = \frac{1}{-9} + \frac{1}{u} \] \[ \frac{1}{u} = \frac{-1+2}{18} \] \[ u = +18\,cm \]
12. Given: \( m = 0.5 \), \( v = +16\,cm \) \[ 0.5 = -\frac{16}{u} \] \[ u = -32\,cm \] \[ \frac{1}{f} = \frac{1}{16} + \frac{1}{-32} \] \[ f = +32\,cm \]
13. Object at focus \[ v = \infty \]
14. Equal size real image \[ u = 2f \]
15. Given: \( u = -20\,cm \), \( f = +15\,cm \) \[ \frac{1}{15} = \frac{1}{v} + \frac{1}{-20} \] \[ v = 60\,cm \]
16. Given: \( m = -3 \), \( v = -30\,cm \) \[ -3 = -\frac{-30}{u} \] \[ u = -10\,cm \]
17. Given: \( u=-24\,cm \), \( v=+12\,cm \) \[ \frac{1}{f} = \frac{1}{12} + \frac{1}{-24} \] \[ f = +24\,cm \]
18. Plane mirror Object = -8 cm Image = +8 cm
19. \( f = -25\,cm \)
20. Given: \( v = +15\,cm \) \[ m = -\frac{15}{u} \] \[ m > 0 \]
21. Given: \( f=-10\,cm \), \( v=-20\,cm \) \[ \frac{1}{-10} = \frac{1}{-20} + \frac{1}{u} \] \[ u = -20\,cm \]
22. Given: \( f=+30\,cm \), \( v=+10\,cm \) \[ \frac{1}{30} = \frac{1}{10} + \frac{1}{u} \] \[ u = -15\,cm \]
23. Plane mirror Image = +15 cm \[ m = +1 \]
24. Given: \( u=-20\,cm \), \( v=-40\,cm \) \[ \frac{1}{f} = \frac{1}{-40} + \frac{1}{-20} \] \[ f = -13.3\,cm \]
25. Given: \( m = \frac{2}{8} = 0.25 \) \[ 0.25 = -\frac{v}{u} \] Solve to get: \[ f = +32\,cm \]
Solution – Concave Mirror (Object Between Pole and Focus)
Given:
Focal length = 12 cm (Concave mirror)
Object distance = 8 cm (between pole and focus)
Step 1: Apply Sign Convention For concave mirror: \[ f = -12\,cm \] Object placed in front of mirror: \[ u = -8\,cm \]
Step 2: Use Mirror Formula \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] Substitute values: \[ \frac{1}{-12} = \frac{1}{v} + \frac{1}{-8} \] \[ \frac{1}{-12} = \frac{1}{v} – \frac{1}{8} \]
Step 3: Solve for v \[ \frac{1}{v} = \frac{1}{-12} + \frac{1}{8} \] Take LCM = 24 \[ \frac{1}{v} = \frac{-2 + 3}{24} \] \[ \frac{1}{v} = \frac{1}{24} \] \[ v = +24\,cm \]
Step 4: Calculate Magnification \[ m = -\frac{v}{u} \] \[ m = -\frac{24}{-8} \] \[ m = +3 \]
Final Answer:
Image distance: \[ v = +24\,cm \] Magnification: \[ m = +3 \]
Nature of Image:
• Image is virtual (since v is positive).
• Image is erect (since m is positive).
• Image is magnified (three times).
Focal length = 12 cm (Concave mirror)
Object distance = 8 cm (between pole and focus)
Step 1: Apply Sign Convention For concave mirror: \[ f = -12\,cm \] Object placed in front of mirror: \[ u = -8\,cm \]
Step 2: Use Mirror Formula \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] Substitute values: \[ \frac{1}{-12} = \frac{1}{v} + \frac{1}{-8} \] \[ \frac{1}{-12} = \frac{1}{v} – \frac{1}{8} \]
Step 3: Solve for v \[ \frac{1}{v} = \frac{1}{-12} + \frac{1}{8} \] Take LCM = 24 \[ \frac{1}{v} = \frac{-2 + 3}{24} \] \[ \frac{1}{v} = \frac{1}{24} \] \[ v = +24\,cm \]
Step 4: Calculate Magnification \[ m = -\frac{v}{u} \] \[ m = -\frac{24}{-8} \] \[ m = +3 \]
Final Answer:
Image distance: \[ v = +24\,cm \] Magnification: \[ m = +3 \]
Nature of Image:
• Image is virtual (since v is positive).
• Image is erect (since m is positive).
• Image is magnified (three times).