Class 10 Physics – Light
CBSE Board Standard Level Numericals (Sign Convention Practice)
Important Formulae
- Mirror Formula: 1/f = 1/v + 1/u
- Lens Formula: 1/f = 1/v – 1/u
- Magnification (Mirror): m = –v/u
- Magnification (Lens): m = v/u
- Power of Lens: P = 1/f (f in metre)
Part A – Mirrors (Board Pattern)
- A concave mirror of focal length 15 cm forms an image at 30 cm. Find the object distance and nature of image.
- An object is placed 25 cm from a concave mirror of focal length 10 cm. Find the image distance.
- A convex mirror of focal length 20 cm forms an image at 8 cm. Find the object distance and magnification.
- A concave mirror of radius of curvature 30 cm forms an image 10 cm in front of the mirror. Find the object distance.
- An object is placed 18 cm from a concave mirror and forms an image at 9 cm. Find the focal length.
- A convex mirror produces an image of height 3 cm for an object of height 6 cm. If object distance is 24 cm, find the focal length.
- An object is placed 40 cm from a concave mirror and a real image forms at 10 cm. Find the focal length.
- Find the position of image formed by a convex mirror of focal length 25 cm when object is at 50 cm.
- An object of height 5 cm is placed 20 cm in front of a convex mirror of focal length 15 cm. Determine image height.
- An object is placed 12 cm from a concave mirror of focal length 8 cm. Find image distance and magnification.
Part B – Lenses (Board Pattern)
- A convex lens of focal length 20 cm forms image of object placed at 40 cm. Find image distance.
- An object is placed 30 cm from a convex lens of focal length 10 cm. Locate image and state nature.
- An image is formed at 30 cm by a convex lens of focal length 15 cm. Find object distance.
- A concave lens of focal length 12 cm forms image at 8 cm. Find object distance.
- A convex lens of focal length 25 cm produces magnification 2. If image distance is 50 cm, find object distance.
- An object of height 4 cm is placed at 15 cm from convex lens of focal length 10 cm. Find image height.
- A convex lens forms virtual image at 10 cm when object is at 15 cm. Determine focal length.
- A concave lens of focal length 20 cm produces image at 16 cm. Find object distance.
- An object of height 8 cm is placed 30 cm in front of convex lens of focal length 15 cm. Find size of image.
- A convex lens of focal length 30 cm forms real image at 60 cm. Find object distance and magnification.
Part C – Mixed Standard Questions
- A concave mirror of focal length 18 cm forms image at 6 cm when object is placed at 9 cm. Find magnification.
- An object is placed 45 cm in front of convex mirror of focal length 15 cm. Find image distance.
- A convex lens forms real and inverted image at 25 cm when object is at 15 cm. Find focal length.
- A concave mirror has radius of curvature 40 cm. Object placed at 12 cm. Determine image distance and magnification.
- An object of height 6 cm is placed at 20 cm from convex lens of focal length 10 cm. Find image distance and size of image.
Note on Sign Convention (Students Must Apply Themselves)
- Apply New Cartesian Sign Convention.
- Decide the sign of focal length according to mirror or lens type.
- Object distance is taken as per direction of incident light.
- Final answer must include nature of image wherever required.
Class 10 Physics – Light
CBSE Board Standard Level Numericals (Sign Convention)
Important Formulae
- Mirror Formula: 1/f = 1/v + 1/u
- Lens Formula: 1/f = 1/v – 1/u
- Magnification (Mirror): m = –v/u
- Magnification (Lens): m = v/u
- Power of Lens: P = 1/f (f in metre)
Part A – Mirrors (Board Pattern)
- A concave mirror of focal length –15 cm forms an image at –30 cm. Find the object distance and nature of image.
- An object is placed at –25 cm from a concave mirror of focal length –10 cm. Find the image distance.
- A convex mirror of focal length +20 cm forms an image at +8 cm. Find the object distance and magnification.
- A concave mirror of radius of curvature –30 cm forms an image 10 cm in front of the mirror. Find the object distance.
- An object is placed at –18 cm from a concave mirror and forms an image at –9 cm. Find the focal length.
- A convex mirror produces an image of height 3 cm for an object of height 6 cm. If object distance is –24 cm, find the focal length.
- An object is placed at –40 cm from a concave mirror and a real image forms at –10 cm. Find the focal length.
- Find the position of image formed by a convex mirror of focal length +25 cm when object is at –50 cm.
- An object of height 5 cm is placed 20 cm in front of a convex mirror of focal length +15 cm. Determine image height.
- An object is placed at –12 cm from a concave mirror of focal length –8 cm. Find image distance and magnification.
Part B – Lenses (Board Pattern)
- A convex lens of focal length +20 cm forms image of object at –40 cm. Find image distance.
- An object is placed at –30 cm from a convex lens of focal length +10 cm. Locate image and state nature.
- An image is formed at +30 cm by a convex lens of focal length +15 cm. Find object distance.
- A concave lens of focal length –12 cm forms image at –8 cm. Find object distance.
- A convex lens of focal length +25 cm produces magnification –2. If image distance is +50 cm, find object distance.
- An object of height 4 cm is placed at –15 cm from convex lens of focal length +10 cm. Find image height.
- A convex lens forms virtual image at –10 cm when object is at –15 cm. Determine focal length.
- A concave lens of focal length –20 cm produces image at –16 cm. Find object distance.
- An object of height 8 cm is placed 30 cm in front of convex lens of focal length +15 cm. Find size of image.
- A convex lens of focal length +30 cm forms real image at +60 cm. Find object distance and magnification.
Part C – Mixed Standard Questions
- A concave mirror of focal length –18 cm forms image at –6 cm when object is placed at –9 cm. Find magnification.
- An object is placed at –45 cm in front of convex mirror of focal length +15 cm. Find image distance.
- A convex lens forms real and inverted image at +25 cm when object is at –15 cm. Find focal length.
- A concave mirror has radius of curvature –40 cm. Object placed at –12 cm. Determine image distance and magnification.
- An object of height 6 cm is placed at –20 cm from convex lens of focal length +10 cm. Find image distance and size of image.
Note on Sign Convention (New Cartesian Convention)
- All distances are measured from Pole (mirror) or Optical Centre (lens).
- Distances measured in direction of incident light are positive.
- Distances measured opposite to incident light are negative.
- Focal length of concave mirror is negative; convex mirror is positive.
- Focal length of convex lens is positive; concave lens is negative.
Part A – Mirrors (Detailed Solutions)
1. Concave Mirror
Given: \( f = -15\,cm \), \( v = -30\,cm \) Mirror Formula: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] \[ \frac{1}{-15} = \frac{1}{-30} + \frac{1}{u} \] \[ \frac{1}{u} = \frac{1}{-15} – \frac{1}{-30} \] LCM = 30 \[ \frac{1}{u} = \frac{-2+1}{30} = \frac{-1}{30} \] \[ u = -30\,cm \] Object at 30 cm in front. Image is real and inverted.
2. Concave Mirror
Given: \( f=-10\,cm \), \( u=-25\,cm \) \[ \frac{1}{-10} = \frac{1}{v} + \frac{1}{-25} \] \[ \frac{1}{v} = \frac{1}{-10} – \frac{1}{-25} \] LCM = 50 \[ \frac{1}{v} = \frac{-5+2}{50} = \frac{-3}{50} \] \[ v = -16.7\,cm \] Real and inverted.
3. Convex Mirror
Given: \( f=+20\,cm \), \( v=+8\,cm \) \[ \frac{1}{20} = \frac{1}{8} + \frac{1}{u} \] \[ \frac{1}{u} = \frac{1}{20} – \frac{1}{8} \] LCM = 40 \[ \frac{1}{u} = \frac{2-5}{40} = \frac{-3}{40} \] \[ u = -13.3\,cm \] Magnification: \[ m = -\frac{v}{u} = 0.6 \] Virtual and erect.
4. Concave Mirror
Radius \( R = -30\,cm \) \[ f = \frac{R}{2} = -15\,cm \] Given \( v = -10\,cm \) \[ \frac{1}{-15} = \frac{1}{-10} + \frac{1}{u} \] \[ \frac{1}{u} = \frac{-2+3}{30} \] \[ u = +30\,cm \]
5. Concave Mirror
Given \( u=-18\,cm \), \( v=-9\,cm \) \[ \frac{1}{f} = \frac{1}{-9} + \frac{1}{-18} \] \[ \frac{1}{f} = \frac{-2-1}{18} = \frac{-3}{18} \] \[ f = -6\,cm \]
6. Convex Mirror
Given: \( h_o = 6\,cm \), \( h_i = 3\,cm \), \( u=-24\,cm \) \[ m = \frac{h_i}{h_o} = \frac{3}{6} = 0.5 \] \[ m = -\frac{v}{u} \] \[ 0.5 = -\frac{v}{-24} \] \[ v = +12\,cm \] \[ \frac{1}{f} = \frac{1}{12} + \frac{1}{-24} \] \[ \frac{1}{f} = \frac{2-1}{24} = \frac{1}{24} \] \[ f = +24\,cm \]
7. Concave Mirror
Given \( u=-40\,cm \), \( v=-10\,cm \) \[ \frac{1}{f} = \frac{1}{-10} + \frac{1}{-40} \] \[ \frac{1}{f} = \frac{-4-1}{40} = \frac{-5}{40} \] \[ f = -8\,cm \]
8. Convex Mirror
Given \( f=+25\,cm \), \( u=-50\,cm \) \[ \frac{1}{25} = \frac{1}{v} + \frac{1}{-50} \] \[ \frac{1}{v} = \frac{1}{25} + \frac{1}{50} \] \[ \frac{1}{v} = \frac{2+1}{50} = \frac{3}{50} \] \[ v = 16.7\,cm \]
9. Convex Mirror
\[ \frac{1}{15} = \frac{1}{v} + \frac{1}{-20} \] \[ \frac{1}{v} = \frac{1}{15} + \frac{1}{20} \] \[ \frac{1}{v} = \frac{4+3}{60} = \frac{7}{60} \] \[ v = 8.57\,cm \] \[ m = -\frac{v}{u} = 0.43 \] \[ h_i = 0.43 \times 5 = 2.15\,cm \]
10. Concave Mirror
\[ \frac{1}{-8} = \frac{1}{v} + \frac{1}{-12} \] \[ \frac{1}{v} = \frac{-3+2}{24} \] \[ v = -24\,cm \] \[ m = -\frac{-24}{-12} = -2 \] Real and inverted.
Given: \( f = -15\,cm \), \( v = -30\,cm \) Mirror Formula: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] \[ \frac{1}{-15} = \frac{1}{-30} + \frac{1}{u} \] \[ \frac{1}{u} = \frac{1}{-15} – \frac{1}{-30} \] LCM = 30 \[ \frac{1}{u} = \frac{-2+1}{30} = \frac{-1}{30} \] \[ u = -30\,cm \] Object at 30 cm in front. Image is real and inverted.
2. Concave Mirror
Given: \( f=-10\,cm \), \( u=-25\,cm \) \[ \frac{1}{-10} = \frac{1}{v} + \frac{1}{-25} \] \[ \frac{1}{v} = \frac{1}{-10} – \frac{1}{-25} \] LCM = 50 \[ \frac{1}{v} = \frac{-5+2}{50} = \frac{-3}{50} \] \[ v = -16.7\,cm \] Real and inverted.
3. Convex Mirror
Given: \( f=+20\,cm \), \( v=+8\,cm \) \[ \frac{1}{20} = \frac{1}{8} + \frac{1}{u} \] \[ \frac{1}{u} = \frac{1}{20} – \frac{1}{8} \] LCM = 40 \[ \frac{1}{u} = \frac{2-5}{40} = \frac{-3}{40} \] \[ u = -13.3\,cm \] Magnification: \[ m = -\frac{v}{u} = 0.6 \] Virtual and erect.
4. Concave Mirror
Radius \( R = -30\,cm \) \[ f = \frac{R}{2} = -15\,cm \] Given \( v = -10\,cm \) \[ \frac{1}{-15} = \frac{1}{-10} + \frac{1}{u} \] \[ \frac{1}{u} = \frac{-2+3}{30} \] \[ u = +30\,cm \]
5. Concave Mirror
Given \( u=-18\,cm \), \( v=-9\,cm \) \[ \frac{1}{f} = \frac{1}{-9} + \frac{1}{-18} \] \[ \frac{1}{f} = \frac{-2-1}{18} = \frac{-3}{18} \] \[ f = -6\,cm \]
6. Convex Mirror
Given: \( h_o = 6\,cm \), \( h_i = 3\,cm \), \( u=-24\,cm \) \[ m = \frac{h_i}{h_o} = \frac{3}{6} = 0.5 \] \[ m = -\frac{v}{u} \] \[ 0.5 = -\frac{v}{-24} \] \[ v = +12\,cm \] \[ \frac{1}{f} = \frac{1}{12} + \frac{1}{-24} \] \[ \frac{1}{f} = \frac{2-1}{24} = \frac{1}{24} \] \[ f = +24\,cm \]
7. Concave Mirror
Given \( u=-40\,cm \), \( v=-10\,cm \) \[ \frac{1}{f} = \frac{1}{-10} + \frac{1}{-40} \] \[ \frac{1}{f} = \frac{-4-1}{40} = \frac{-5}{40} \] \[ f = -8\,cm \]
8. Convex Mirror
Given \( f=+25\,cm \), \( u=-50\,cm \) \[ \frac{1}{25} = \frac{1}{v} + \frac{1}{-50} \] \[ \frac{1}{v} = \frac{1}{25} + \frac{1}{50} \] \[ \frac{1}{v} = \frac{2+1}{50} = \frac{3}{50} \] \[ v = 16.7\,cm \]
9. Convex Mirror
\[ \frac{1}{15} = \frac{1}{v} + \frac{1}{-20} \] \[ \frac{1}{v} = \frac{1}{15} + \frac{1}{20} \] \[ \frac{1}{v} = \frac{4+3}{60} = \frac{7}{60} \] \[ v = 8.57\,cm \] \[ m = -\frac{v}{u} = 0.43 \] \[ h_i = 0.43 \times 5 = 2.15\,cm \]
10. Concave Mirror
\[ \frac{1}{-8} = \frac{1}{v} + \frac{1}{-12} \] \[ \frac{1}{v} = \frac{-3+2}{24} \] \[ v = -24\,cm \] \[ m = -\frac{-24}{-12} = -2 \] Real and inverted.
Part B – Lenses (Detailed Solutions)
11. Convex Lens
Given: \( f = +20\,cm \), \( u = -40\,cm \) Lens Formula: \[ \frac{1}{f} = \frac{1}{v} – \frac{1}{u} \] \[ \frac{1}{20} = \frac{1}{v} – \left(\frac{1}{-40}\right) \] \[ \frac{1}{20} = \frac{1}{v} + \frac{1}{40} \] \[ \frac{1}{v} = \frac{1}{20} – \frac{1}{40} \] LCM = 40 \[ \frac{1}{v} = \frac{2 – 1}{40} = \frac{1}{40} \] \[ v = +40\,cm \] Image real and inverted.
12. Convex Lens
Given: \( f = +10\,cm \), \( u = -30\,cm \) \[ \frac{1}{10} = \frac{1}{v} – \left(\frac{1}{-30}\right) \] \[ \frac{1}{10} = \frac{1}{v} + \frac{1}{30} \] \[ \frac{1}{v} = \frac{1}{10} – \frac{1}{30} \] LCM = 30 \[ \frac{1}{v} = \frac{3 – 1}{30} = \frac{2}{30} \] \[ v = +15\,cm \] Image real and inverted.
13. Convex Lens
Given: \( f = +15\,cm \), \( v = +30\,cm \) \[ \frac{1}{15} = \frac{1}{30} – \frac{1}{u} \] \[ \frac{1}{u} = \frac{1}{30} – \frac{1}{15} \] LCM = 30 \[ \frac{1}{u} = \frac{1 – 2}{30} = \frac{-1}{30} \] \[ u = -30\,cm \]
14. Concave Lens
Given: \( f = -12\,cm \), \( v = -8\,cm \) \[ \frac{1}{-12} = \frac{1}{-8} – \frac{1}{u} \] \[ \frac{1}{u} = \frac{1}{-8} – \frac{1}{-12} \] LCM = 24 \[ \frac{1}{u} = \frac{-3 + 2}{24} = \frac{-1}{24} \] \[ u = -24\,cm \]
15. Convex Lens
Given: \( f = +25\,cm \), \( v = +50\,cm \) \[ \frac{1}{25} = \frac{1}{50} – \frac{1}{u} \] \[ \frac{1}{u} = \frac{1}{50} – \frac{1}{25} \] LCM = 50 \[ \frac{1}{u} = \frac{1 – 2}{50} = \frac{-1}{50} \] \[ u = -50\,cm \]
16. Convex Lens
Given: \( h_o = 4\,cm \), \( u = -15\,cm \), \( f = +10\,cm \) \[ \frac{1}{10} = \frac{1}{v} – \left(\frac{1}{-15}\right) \] \[ \frac{1}{10} = \frac{1}{v} + \frac{1}{15} \] \[ \frac{1}{v} = \frac{1}{10} – \frac{1}{15} \] LCM = 30 \[ \frac{1}{v} = \frac{3 – 2}{30} = \frac{1}{30} \] \[ v = +30\,cm \] Magnification: \[ m = \frac{v}{u} = \frac{30}{-15} = -2 \] \[ h_i = -2 \times 4 = -8\,cm \] Image height = 8 cm (real and inverted).
17. Convex Lens
Given: \( u = -15\,cm \), \( v = -10\,cm \) \[ \frac{1}{f} = \frac{1}{-10} – \frac{1}{-15} \] LCM = 30 \[ \frac{1}{f} = \frac{-3 + 2}{30} = \frac{-1}{30} \] \[ f = -30\,cm \]
18. Concave Lens
Given: \( f = -20\,cm \), \( v = -16\,cm \) \[ \frac{1}{-20} = \frac{1}{-16} – \frac{1}{u} \] LCM = 80 \[ \frac{1}{u} = \frac{-5 + 4}{80} = \frac{-1}{80} \] \[ u = -80\,cm \]
19. Convex Lens
Given: \( u = -30\,cm \), \( f = +15\,cm \) \[ \frac{1}{15} = \frac{1}{v} – \left(\frac{1}{-30}\right) \] \[ \frac{1}{15} = \frac{1}{v} + \frac{1}{30} \] \[ \frac{1}{v} = \frac{1}{15} – \frac{1}{30} \] LCM = 30 \[ \frac{1}{v} = \frac{2 – 1}{30} = \frac{1}{30} \] \[ v = +30\,cm \] Magnification: \[ m = \frac{v}{u} = \frac{30}{-30} = -1 \] \[ h_i = -1 \times 8 = -8\,cm \]
20. Convex Lens
Given: \( f = +30\,cm \), \( v = +60\,cm \) \[ \frac{1}{30} = \frac{1}{60} – \frac{1}{u} \] \[ \frac{1}{u} = \frac{1}{60} – \frac{1}{30} \] LCM = 60 \[ \frac{1}{u} = \frac{1 – 2}{60} = \frac{-1}{60} \] \[ u = -60\,cm \] Magnification: \[ m = \frac{v}{u} = \frac{60}{-60} = -1 \]
Given: \( f = +20\,cm \), \( u = -40\,cm \) Lens Formula: \[ \frac{1}{f} = \frac{1}{v} – \frac{1}{u} \] \[ \frac{1}{20} = \frac{1}{v} – \left(\frac{1}{-40}\right) \] \[ \frac{1}{20} = \frac{1}{v} + \frac{1}{40} \] \[ \frac{1}{v} = \frac{1}{20} – \frac{1}{40} \] LCM = 40 \[ \frac{1}{v} = \frac{2 – 1}{40} = \frac{1}{40} \] \[ v = +40\,cm \] Image real and inverted.
12. Convex Lens
Given: \( f = +10\,cm \), \( u = -30\,cm \) \[ \frac{1}{10} = \frac{1}{v} – \left(\frac{1}{-30}\right) \] \[ \frac{1}{10} = \frac{1}{v} + \frac{1}{30} \] \[ \frac{1}{v} = \frac{1}{10} – \frac{1}{30} \] LCM = 30 \[ \frac{1}{v} = \frac{3 – 1}{30} = \frac{2}{30} \] \[ v = +15\,cm \] Image real and inverted.
13. Convex Lens
Given: \( f = +15\,cm \), \( v = +30\,cm \) \[ \frac{1}{15} = \frac{1}{30} – \frac{1}{u} \] \[ \frac{1}{u} = \frac{1}{30} – \frac{1}{15} \] LCM = 30 \[ \frac{1}{u} = \frac{1 – 2}{30} = \frac{-1}{30} \] \[ u = -30\,cm \]
14. Concave Lens
Given: \( f = -12\,cm \), \( v = -8\,cm \) \[ \frac{1}{-12} = \frac{1}{-8} – \frac{1}{u} \] \[ \frac{1}{u} = \frac{1}{-8} – \frac{1}{-12} \] LCM = 24 \[ \frac{1}{u} = \frac{-3 + 2}{24} = \frac{-1}{24} \] \[ u = -24\,cm \]
15. Convex Lens
Given: \( f = +25\,cm \), \( v = +50\,cm \) \[ \frac{1}{25} = \frac{1}{50} – \frac{1}{u} \] \[ \frac{1}{u} = \frac{1}{50} – \frac{1}{25} \] LCM = 50 \[ \frac{1}{u} = \frac{1 – 2}{50} = \frac{-1}{50} \] \[ u = -50\,cm \]
16. Convex Lens
Given: \( h_o = 4\,cm \), \( u = -15\,cm \), \( f = +10\,cm \) \[ \frac{1}{10} = \frac{1}{v} – \left(\frac{1}{-15}\right) \] \[ \frac{1}{10} = \frac{1}{v} + \frac{1}{15} \] \[ \frac{1}{v} = \frac{1}{10} – \frac{1}{15} \] LCM = 30 \[ \frac{1}{v} = \frac{3 – 2}{30} = \frac{1}{30} \] \[ v = +30\,cm \] Magnification: \[ m = \frac{v}{u} = \frac{30}{-15} = -2 \] \[ h_i = -2 \times 4 = -8\,cm \] Image height = 8 cm (real and inverted).
17. Convex Lens
Given: \( u = -15\,cm \), \( v = -10\,cm \) \[ \frac{1}{f} = \frac{1}{-10} – \frac{1}{-15} \] LCM = 30 \[ \frac{1}{f} = \frac{-3 + 2}{30} = \frac{-1}{30} \] \[ f = -30\,cm \]
18. Concave Lens
Given: \( f = -20\,cm \), \( v = -16\,cm \) \[ \frac{1}{-20} = \frac{1}{-16} – \frac{1}{u} \] LCM = 80 \[ \frac{1}{u} = \frac{-5 + 4}{80} = \frac{-1}{80} \] \[ u = -80\,cm \]
19. Convex Lens
Given: \( u = -30\,cm \), \( f = +15\,cm \) \[ \frac{1}{15} = \frac{1}{v} – \left(\frac{1}{-30}\right) \] \[ \frac{1}{15} = \frac{1}{v} + \frac{1}{30} \] \[ \frac{1}{v} = \frac{1}{15} – \frac{1}{30} \] LCM = 30 \[ \frac{1}{v} = \frac{2 – 1}{30} = \frac{1}{30} \] \[ v = +30\,cm \] Magnification: \[ m = \frac{v}{u} = \frac{30}{-30} = -1 \] \[ h_i = -1 \times 8 = -8\,cm \]
20. Convex Lens
Given: \( f = +30\,cm \), \( v = +60\,cm \) \[ \frac{1}{30} = \frac{1}{60} – \frac{1}{u} \] \[ \frac{1}{u} = \frac{1}{60} – \frac{1}{30} \] LCM = 60 \[ \frac{1}{u} = \frac{1 – 2}{60} = \frac{-1}{60} \] \[ u = -60\,cm \] Magnification: \[ m = \frac{v}{u} = \frac{60}{-60} = -1 \]
Part C – Mixed Standard Questions (Detailed Solutions)
21. Concave Mirror
Given: \( f = -18\,cm \), \( v = -6\,cm \) Mirror formula: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] \[ \frac{1}{-18} = \frac{1}{-6} + \frac{1}{u} \] \[ \frac{1}{u} = \frac{1}{-18} – \frac{1}{-6} \] LCM = 18 \[ \frac{1}{u} = \frac{-1 + 3}{18} = \frac{2}{18} \] \[ u = +9\,cm \] Magnification: \[ m = -\frac{v}{u} = -\frac{-6}{9} = 0.67 \]
22. Convex Mirror
Given: \( f = +15\,cm \), \( u = -45\,cm \) \[ \frac{1}{15} = \frac{1}{v} + \frac{1}{-45} \] \[ \frac{1}{v} = \frac{1}{15} + \frac{1}{45} \] LCM = 45 \[ \frac{1}{v} = \frac{3 + 1}{45} = \frac{4}{45} \] \[ v = 11.25\,cm \] Image formed behind mirror (virtual and erect).
23. Convex Lens
Given: \( v = +25\,cm \), \( u = -15\,cm \) Lens formula: \[ \frac{1}{f} = \frac{1}{v} – \frac{1}{u} \] \[ \frac{1}{f} = \frac{1}{25} – \left(\frac{1}{-15}\right) \] \[ \frac{1}{f} = \frac{1}{25} + \frac{1}{15} \] LCM = 75 \[ \frac{1}{f} = \frac{3 + 5}{75} = \frac{8}{75} \] \[ f = 9.4\,cm \]
24. Concave Mirror
Radius of curvature \( R = -40\,cm \) \[ f = \frac{R}{2} = -20\,cm \] Given \( u = -12\,cm \) \[ \frac{1}{-20} = \frac{1}{v} + \frac{1}{-12} \] \[ \frac{1}{v} = \frac{1}{-20} – \frac{1}{-12} \] LCM = 60 \[ \frac{1}{v} = \frac{-3 + 5}{60} = \frac{2}{60} \] \[ v = 30\,cm \] Magnification: \[ m = -\frac{v}{u} = -\frac{30}{-12} = 2.5 \]
25. Convex Lens
Given: \( f = +10\,cm \), \( u = -20\,cm \) \[ \frac{1}{10} = \frac{1}{v} – \left(\frac{1}{-20}\right) \] \[ \frac{1}{10} = \frac{1}{v} + \frac{1}{20} \] \[ \frac{1}{v} = \frac{1}{10} – \frac{1}{20} \] LCM = 20 \[ \frac{1}{v} = \frac{2 – 1}{20} = \frac{1}{20} \] \[ v = 20\,cm \] Magnification: \[ m = \frac{v}{u} = \frac{20}{-20} = -1 \] Image height: \[ h_i = -1 \times 6 = -6\,cm \] Image height = 6 cm (real and inverted).
Given: \( f = -18\,cm \), \( v = -6\,cm \) Mirror formula: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] \[ \frac{1}{-18} = \frac{1}{-6} + \frac{1}{u} \] \[ \frac{1}{u} = \frac{1}{-18} – \frac{1}{-6} \] LCM = 18 \[ \frac{1}{u} = \frac{-1 + 3}{18} = \frac{2}{18} \] \[ u = +9\,cm \] Magnification: \[ m = -\frac{v}{u} = -\frac{-6}{9} = 0.67 \]
22. Convex Mirror
Given: \( f = +15\,cm \), \( u = -45\,cm \) \[ \frac{1}{15} = \frac{1}{v} + \frac{1}{-45} \] \[ \frac{1}{v} = \frac{1}{15} + \frac{1}{45} \] LCM = 45 \[ \frac{1}{v} = \frac{3 + 1}{45} = \frac{4}{45} \] \[ v = 11.25\,cm \] Image formed behind mirror (virtual and erect).
23. Convex Lens
Given: \( v = +25\,cm \), \( u = -15\,cm \) Lens formula: \[ \frac{1}{f} = \frac{1}{v} – \frac{1}{u} \] \[ \frac{1}{f} = \frac{1}{25} – \left(\frac{1}{-15}\right) \] \[ \frac{1}{f} = \frac{1}{25} + \frac{1}{15} \] LCM = 75 \[ \frac{1}{f} = \frac{3 + 5}{75} = \frac{8}{75} \] \[ f = 9.4\,cm \]
24. Concave Mirror
Radius of curvature \( R = -40\,cm \) \[ f = \frac{R}{2} = -20\,cm \] Given \( u = -12\,cm \) \[ \frac{1}{-20} = \frac{1}{v} + \frac{1}{-12} \] \[ \frac{1}{v} = \frac{1}{-20} – \frac{1}{-12} \] LCM = 60 \[ \frac{1}{v} = \frac{-3 + 5}{60} = \frac{2}{60} \] \[ v = 30\,cm \] Magnification: \[ m = -\frac{v}{u} = -\frac{30}{-12} = 2.5 \]
25. Convex Lens
Given: \( f = +10\,cm \), \( u = -20\,cm \) \[ \frac{1}{10} = \frac{1}{v} – \left(\frac{1}{-20}\right) \] \[ \frac{1}{10} = \frac{1}{v} + \frac{1}{20} \] \[ \frac{1}{v} = \frac{1}{10} – \frac{1}{20} \] LCM = 20 \[ \frac{1}{v} = \frac{2 – 1}{20} = \frac{1}{20} \] \[ v = 20\,cm \] Magnification: \[ m = \frac{v}{u} = \frac{20}{-20} = -1 \] Image height: \[ h_i = -1 \times 6 = -6\,cm \] Image height = 6 cm (real and inverted).