Most Important Numericals of Mirror Formula of Light Chapter Class 10th With Explanation that are beneficial for the various board exams.
50 most Important Mirror Formula numerical problems Master Class
50 Professionally Verified Numerical Problems — Class 10
Q1. An object is placed at a distance of 15 cm from a concave mirror of focal length 10 cm. Find the position and nature of the image.
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Given: $u = -15\text{ cm}$, $f = -10\text{ cm}$.
Formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} = \frac{1}{-10} – \frac{1}{-15} = \frac{-3 + 2}{30} = -\frac{1}{30}$
$\therefore v = -30\text{ cm}$.
Nature: Real, inverted, and magnified.
Q2. A 2 cm high object is placed 20 cm in front of a concave mirror of focal length 15 cm. Find the position and size of the image.
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Given: $u = -20\text{ cm}$, $f = -15\text{ cm}$, $h_o = +2\text{ cm}$.
$\frac{1}{v} = \frac{1}{-15} – \frac{1}{-20} = -\frac{1}{60} \implies v = -60\text{ cm}$.
$m = -\frac{v}{u} = -\frac{-60}{-20} = -3$.
$h_i = m \times h_o = -3 \times 2 = -6\text{ cm}$.
Nature: Real, inverted, 60 cm in front of the mirror.
Q3. An object is kept 5 cm in front of a concave mirror of focal length 10 cm. Calculate the image position.
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Given: $u = -5\text{ cm}$, $f = -10\text{ cm}$.
$\frac{1}{v} = \frac{1}{-10} – \frac{1}{-5} = \frac{1}{10} \implies v = +10\text{ cm}$.
Nature: Virtual, erect, formed 10 cm behind the mirror.
Q4. Find the position of the image when an object is placed 30 cm from a concave mirror of focal length 15 cm.
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Given: $u = -30\text{ cm}$, $f = -15\text{ cm}$.
$\frac{1}{v} = \frac{1}{-15} – \frac{1}{-30} = -\frac{1}{30} \implies v = -30\text{ cm}$.
Image forms at the center of curvature ($C$). Real and inverted.
Q5. At what distance from a concave mirror of focal length 25 cm should an object be placed so that the image is formed at infinity?
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To form an image at infinity, the object must be at the principal focus. By sign convention, $u = -25\text{ cm}$.
Q6. An object is placed at 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
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Given: $u = -10\text{ cm}$, $f = +15\text{ cm}$.
$\frac{1}{v} = \frac{1}{15} – \frac{1}{-10} = \frac{5}{30} = \frac{1}{6} \implies v = +6\text{ cm}$.
Nature: Virtual, erect, and diminished.
Q7. A convex mirror used for rear-view has a radius of curvature of 3.00 m. If a bus is 5.00 m from this mirror, find the image position.
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Given: $R = +3.00\text{ m} \implies f = +1.50\text{ m}$, $u = -5.00\text{ m}$.
$\frac{1}{v} = \frac{1}{1.50} – \frac{1}{-5.00} = \frac{13}{15} \implies v = +1.15\text{ m}$.
Nature: Virtual, erect, formed 1.15 m behind mirror.
Q8. An object 5 cm high is placed 20 cm away from a convex mirror of focal length 10 cm. Find the height of the image.
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Given: $u = -20\text{ cm}$, $f = +10\text{ cm}$, $h_o = +5\text{ cm}$.
$\frac{1}{v} = \frac{1}{10} – \frac{1}{-20} = \frac{3}{20} \implies v = +6.67\text{ cm}$.
$m = -\frac{v}{u} = -\frac{20/3}{-20} = +\frac{1}{3}$.
$h_i = \frac{1}{3} \times 5 = +1.67\text{ cm}$.
Q9. A car is 10 m behind a motorcycle equipped with a convex rearview mirror of focal length 2 m. Where is the car’s image?
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Given: $u = -10\text{ m}$, $f = +2\text{ m}$.
$\frac{1}{v} = \frac{1}{2} – \frac{1}{-10} = \frac{6}{10} \implies v = +1.67\text{ m}$.
Q10. Find the focal length of a convex mirror whose radius of curvature is 32 cm.
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For a convex mirror, $R$ is positive. $R = +32\text{ cm}$.
$f = \frac{R}{2} = \frac{+32}{2} = +16\text{ cm}$.
Q11. A concave mirror produces three times magnified real image of an object placed at 10 cm. Where is the image located?
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Given: Real image $\implies m = -3$, $u = -10\text{ cm}$.
$m = -\frac{v}{u} \implies -3 = -\frac{v}{-10} \implies v = -30\text{ cm}$.
Q12. A concave mirror produces a three times magnified virtual image of an object placed at 10 cm. Find the focal length.
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Given: Virtual $\implies m = +3$, $u = -10\text{ cm}$.
$3 = -\frac{v}{-10} \implies v = +30\text{ cm}$.
$\frac{1}{f} = \frac{1}{30} + \frac{1}{-10} = -\frac{1}{15} \implies f = -15\text{ cm}$.
Q13. The magnification produced by a spherical mirror is -2. What type of mirror is it?
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$m = -2$ means the image is real and inverted. Only a concave mirror forms real, magnified images.
Q14. A convex mirror produces a magnification of 1/2 when the object is placed at 20 cm. Find the focal length.
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Given: $m = +\frac{1}{2}$, $u = -20\text{ cm}$.
$\frac{1}{2} = -\frac{v}{-20} \implies v = +10\text{ cm}$.
$\frac{1}{f} = \frac{1}{10} + \frac{1}{-20} = \frac{1}{20} \implies f = +20\text{ cm}$.
Q15. An object’s real image is formed 4 times its size by a concave mirror. If focal length is 20 cm, find u.
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Given: $f = -20\text{ cm}$, $m = -4$.
$m = \frac{f}{f – u} \implies -4 = \frac{-20}{-20 – u} \implies u = -25\text{ cm}$.
Q16. Where should an object be placed in front of a concave mirror ($f = 20$ cm) to obtain a real image two times magnified?
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Given: $f = -20\text{ cm}$, $m = -2$.
$-2 = \frac{-20}{-20 – u} \implies 40 + 2u = -20 \implies u = -30\text{ cm}$.
Q17. Where should an object be placed in front of a concave mirror ($f = 20$ cm) to obtain a virtual image two times magnified?
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Given: $f = -20\text{ cm}$, $m = +2$.
$2 = \frac{-20}{-20 – u} \implies -40 – 2u = -20 \implies u = -10\text{ cm}$.
Q18. An image by a convex mirror is 1/4th the size of the object. If focal length is 30 cm, find object distance.
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Given: $f = +30\text{ cm}$, $m = +\frac{1}{4}$.
$\frac{1}{4} = \frac{30}{30 – u} \implies 30 – u = 120 \implies u = -90\text{ cm}$.
Q19. Object is at 20 cm from a mirror. Image is 15 cm behind the mirror. Find focal length and mirror type.
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Given: $u = -20\text{ cm}$, $v = +15\text{ cm}$.
$\frac{1}{f} = \frac{1}{15} + \frac{1}{-20} = \frac{1}{60} \implies f = +60\text{ cm}$ (Convex).
Q20. A concave mirror forms a real image of an object kept at 10 cm. The image distance is 30 cm. Find the focal length.
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Given: $u = -10\text{ cm}$, $v = -30\text{ cm}$.
$\frac{1}{f} = \frac{1}{-30} + \frac{1}{-10} = -\frac{4}{30} \implies f = -7.5\text{ cm}$.
Q21. The image of a candle flame 30 cm from a mirror is formed on a screen 60 cm in front of it. Find focal length.
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Given: $u = -30\text{ cm}$, $v = -60\text{ cm}$ (Real image on screen).
$\frac{1}{f} = \frac{1}{-60} + \frac{1}{-30} = -\frac{3}{60} \implies f = -20\text{ cm}$.
Q22. A dentist uses a mirror of $f = 3$ cm. She holds it 2 cm from a tooth. What is the magnification?
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Dentist uses a concave mirror. $f = -3\text{ cm}$, $u = -2\text{ cm}$.
$\frac{1}{v} = \frac{1}{-3} – \frac{1}{-2} = \frac{1}{6} \implies v = +6\text{ cm}$.
$m = -\frac{6}{-2} = +3$.
Q23. An object is placed 24 cm from a concave mirror. Image is inverted and double the size. Find focal length.
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Given: $m = -2$ (inverted), $u = -24\text{ cm}$.
$-2 = \frac{f}{f – (-24)} \implies -2f – 48 = f \implies f = -16\text{ cm}$.
Q24. A virtual image, half the size of the object, is formed. The object is 40 cm away. Find the focal length.
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Given: $m = +\frac{1}{2}$ (Convex), $u = -40\text{ cm}$.
$\frac{1}{2} = -\frac{v}{-40} \implies v = +20\text{ cm}$.
$\frac{1}{f} = \frac{1}{20} + \frac{1}{-40} = \frac{1}{40} \implies f = +40\text{ cm}$.
Q25. The radius of curvature of a shaving mirror is 40 cm. Face is 15 cm from it. Find magnification.
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Concave mirror. $f = -20\text{ cm}$, $u = -15\text{ cm}$.
$\frac{1}{v} = \frac{1}{-20} – \frac{1}{-15} = \frac{1}{60} \implies v = +60\text{ cm}$.
$m = -\frac{60}{-15} = +4$.
Q26. Object placed 10 cm from a convex mirror of radius 60 cm. Find image position.
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$u = -10$, $R = +60 \implies f = +30$.
$\frac{1}{v} = \frac{1}{30} – \frac{1}{-10} = \frac{4}{30} \implies v = +7.5\text{ cm}$.
Q27. Concave mirror forms an erect image 3 times the object size. If focal length is 12 cm, find u.
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Given: $m = +3$, $f = -12\text{ cm}$.
$3 = \frac{-12}{-12 – u} \implies -36 – 3u = -12 \implies u = -8\text{ cm}$.
Q28. Find $v$ for an object placed at 40 cm from a concave mirror of $f = 20$ cm.
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$u = -40$, $f = -20$.
$\frac{1}{v} = \frac{1}{-20} – \frac{1}{-40} = -\frac{1}{40} \implies v = -40\text{ cm}$ (At C).
Q29. An object is placed at infinity. Focal length of concave mirror is 15 cm. Where is the image?
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Image is formed at the principal focus. $v = -15\text{ cm}$.
Q30. A convex mirror has $f = 20$ cm. Can it form a real image?
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No. $\frac{1}{v} = \frac{1}{f} – \frac{1}{u}$. Since $f > 0$ and $u < 0$, $\frac{1}{v}$ is always the sum of two positive terms. Thus $v > 0$ always (Virtual).
Q31. Sun’s rays converge 15 cm in front of a concave mirror. Where should an object be placed to get same size image?
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Rays converge at focus $\implies f = -15\text{ cm}$. For same size, object must be at $C$. $R = 2f = -30\text{ cm}$. Object should be at 30 cm ($u = -30\text{ cm}$).
Q32. Object 4 cm in size is placed at 25 cm from a concave mirror of $f = 15$ cm. Find image size.
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$\frac{1}{v} = \frac{1}{-15} – \frac{1}{-25} = -\frac{2}{75} \implies v = -37.5\text{ cm}$.
$m = -\frac{-37.5}{-25} = -1.5 \implies h_i = -1.5 \times 4 = -6\text{ cm}$.
Q33. A mirror forms a virtual image of size 1/3 of the object. If $u = -30$ cm, find $f$.
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$m = +\frac{1}{3}$. $\frac{1}{3} = -\frac{v}{-30} \implies v = +10\text{ cm}$.
$\frac{1}{f} = \frac{1}{10} + \frac{1}{-30} = \frac{2}{30} \implies f = +15\text{ cm}$.
Q34. Calculate magnification when an object is 12 cm from a concave mirror of $f = 8$ cm.
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$\frac{1}{v} = \frac{1}{-8} – \frac{1}{-12} = -\frac{1}{24} \implies v = -24\text{ cm}$.
$m = -\frac{-24}{-12} = -2$.
Q35. Distance required to get a real image magnified 5 times by a concave mirror ($f = -10$ cm)?
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$m = -5$. $-5 = \frac{-10}{-10 – u} \implies 50 + 5u = -10 \implies u = -12\text{ cm}$.
Q36. Convex mirror of focal length 1.5m used for traffic. Truck is at 4.5m. Find image distance.
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$f = +1.5$, $u = -4.5$.
$\frac{1}{v} = \frac{1}{1.5} – \frac{1}{-4.5} = \frac{4}{4.5} \implies v = +1.125\text{ m}$.
Q37. Object is 10 cm from mirror. Image is real, inverted, 20 cm from mirror. Find $f$.
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$u = -10$, $v = -20$.
$\frac{1}{f} = \frac{1}{-20} + \frac{1}{-10} = -\frac{3}{20} \implies f = -6.67\text{ cm}$.
Q38. If magnification is +1, what is the mirror type?
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A plane mirror forms an image with $m = +1$. Spherical mirrors cannot achieve exactly +1 for a real object.
Q39. A spherical mirror produces $m = -1$ on a screen at 50 cm. Find focal length.
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$m = -1 \implies v = u$. Since $v = -50$ (on screen), $u = -50$.
$\frac{1}{f} = \frac{1}{-50} + \frac{1}{-50} = -\frac{2}{50} \implies f = -25\text{ cm}$.
Q40. Object distance is 15 cm. Image is 5 cm behind the mirror. Find $f$.
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$u = -15$, $v = +5$.
$\frac{1}{f} = \frac{1}{5} + \frac{1}{-15} = \frac{2}{15} \implies f = +7.5\text{ cm}$ (Convex).
Q41. Find position of image when object is 8 cm from a concave mirror of $f = 12$ cm.
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$\frac{1}{v} = \frac{1}{-12} – \frac{1}{-8} = \frac{1}{24} \implies v = +24\text{ cm}$.
Q42. A 5 cm object is 10 cm from a concave mirror ($f = 20$ cm). Find image size.
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$\frac{1}{v} = \frac{1}{-20} – \frac{1}{-10} = \frac{1}{20} \implies v = +20\text{ cm}$.
$m = -\frac{20}{-10} = +2 \implies h_i = 5 \times 2 = +10\text{ cm}$.
Q43. Convex mirror $f = 10$ cm. Object at 10 cm. Magnification?
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$u = -10$, $f = +10$. $\frac{1}{v} = \frac{1}{10} – \frac{1}{-10} = \frac{2}{10} \implies v = +5\text{ cm}$.
$m = -\frac{5}{-10} = +0.5$.
Q44. An object is placed at C of a concave mirror. What is its magnification?
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At C, $u = 2f$. Image is at C, so $v = 2f$. $m = -\frac{-2f}{-2f} = -1$.
Q45. The ratio of image height to object height is +0.25. Mirror type?
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$m = +0.25$. Since $m$ is positive and $< 1$, it must be a convex mirror.
Q46. Concave mirror $f = 18$ cm. Find distance of an object for erect image of $m = 3$.
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$m = +3$, $f = -18$. $3 = \frac{-18}{-18 – u} \implies -54 – 3u = -18 \implies u = -12\text{ cm}$.
Q47. Focal length of convex mirror if it forms an image at 10 cm for object at 30 cm?
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$v = +10$ (convex forms virtual), $u = -30$.
$\frac{1}{f} = \frac{1}{10} + \frac{1}{-30} = \frac{2}{30} \implies f = +15\text{ cm}$.
Q48. Concave mirror forms real image at 40 cm for object at 20 cm, calculate m.
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$v = -40$, $u = -20$.
$m = -\frac{-40}{-20} = -2$.
Q49. Where does a convex mirror form an image if an object is moved to infinity?
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At its principal focus behind the mirror ($v = f$).
Q50. A 1 cm object is placed 20 cm from concave mirror ($f = 15$ cm). Find $v$ and $h_i$.
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$\frac{1}{v} = \frac{1}{-15} – \frac{1}{-20} = -\frac{1}{60} \implies v = -60\text{ cm}$.
$m = -\frac{-60}{-20} = -3 \implies h_i = -3 \times 1 = -3\text{ cm}$.