Lens Formula Numerical Class-12 Nootan ISC Physics Solution Ch-16 Refraction of Light at Spherical Surfaces: Lenses. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-12 Physics.
Lens Formula Numerical Class-12 Nootan ISC Physics Solution Ch-16 Refraction of Light at Spherical Surfaces: Lenses
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-16 | Refraction of Light at Spherical Surfaces : Lenses |
| Topics | Numericals on Lens Formula |
| Academic Session | 2025-2026 |
Numericals on Lens Formula
Lens Formula Numerical Class-12 Nootan ISC Physics Solution Ch-16 Refraction of Light at Spherical Surfaces: Lenses
Que-24. An object is placed at 0.2 m from a convex lens of focal length 0.15 m. Find the position of the image.
Ans- m = f / f+u
=> m = 0.15 / 0.15+(-0.2) = v/u
=> 0.15/-0.5 = v / -0.2
=> v = 0.6 m, behind the lens
Que-25. An object is placed at 0.06 m from a convex lens of focal length 0.10 m. Find the position of the image.
Ans- m = f / f+u
=> m = 0.1 / 0.1-0.06) = v/u
=> 0.1/-0.04 = v / -0.06
=> v = -0.15 m
Que-26. A convergent beam of light passes through a diverging lens of focal length 0.2 m and comes to a focus on the axis 0.3 m behind the lens. Where would the beam have been focused in the absence of the lens?
Ans- v = +0.3 , f = -0.2 , u = ?
1/f = 1/v – 1/u
=> 1/-0.2 = 1/0.3 – 1/u
After solving,
=> u = 0.12 m
Que-27. A beam of light converges to a point P. A concave lens of focal length 16 cm is placed in the path of the beam at 12 cm from P. At what point does the beam now converge?
Ans- v = ? , f = -16 , u = +12
1/f = 1/v – 1/u
=> 1/-16 = 1/v – 1/12
After solving,
=> v = +48 cm
Que-28. A double convex lens has 10 cm and 15 cm as its two radii of curvature. The image of an object, placed 30 cm from the lens, is formed at 20 cm from the lens on the other side. Find the focal length and the refractive index of the material of the lens. What will be the focal length of the lens, if it is immersed in water of refractive index 1.33?
Ans- v = +20 , f = ? , u = -30
1/f = 1/v – 1/u
=> 1/f = 1/20 – 1/-30
After solving,
=> f = +12 cm
again 1/f = (1μ2-1) (1/R1 + 1/R2)
=> 1/12 = (1μ2-1) {1/10 + 1/15}
After solving,
=> (1μ2-1) = 6/12
=> 1μ2 = 1.5
again fe/fa = aμg-1 / (aμg/aμe)-1
=> fe/12 = 1.5-1 / (1.5/1.33)-1
=> fe = 47 cm
Que-29. A thin convex lens forms the distinct image of an object on a screen placed at a distance of 30 cm from the lens. When the lens is displaced 5 cm towards the screen, then in order to obtain distinct image the screen is displaced 5 cm towards the lens. Find the focal length of the lens.
Ans-
v = +30 , f = ? , u = -25
1/f = 1/v – 1/u
=> 1/f = 1/30 – 1/-25
After solving,
=> f = +10 cm
— : End of Lens Formula Numerical Class-12 Nootan ISC Physics Solution Ch-16 :–
Return to : – Nootan Solutions for ISC Class-12 Physics
Thanks
Please share with your friends
