Physics · Refraction of light at plane and spherical surfaces, thin lens formula and lens maker formula
A converging beam of rays is incident on a diverging lens. Having passed through
A converging beam of rays is incident on a diverging lens. Having passed through the lens the rays intersect at a point \( 15 \mathrm{cm} \) from the lens on the opposite side. If the lens is removed the point where the rays meet will move 5 cm closer to the lens. The focal length of the lens is :
- A. \( 5 \mathrm{cm} \)
- B. -10 cm
- C. \( 20 \mathrm{cm} \)
- D. -30 cm
Step-by-step solution
The incident converging beam forms a virtual object for the diverging lens. Without the lens, the rays meet at a point 5 cm closer to the lens than with the lens. With lens, the rays meet at 15 cm on the opposite side (real image, v = +15 cm). Without lens, the convergence point is 10 cm from the lens (object distance, u = -10 cm, negative because virtual object). Using the lens formula 1/f = 1/u + 1/v with proper sign convention (u negative for virtual object, v positive for real image, f negative for diverging lens): 1/f = 1/(-10) + 1/15 = -1/10 + 1/15 = -1/30, so f = -30 cm.
Related MCQs
- When an object is placed between optical centre and focus of lens:…
- A needle of length placed from a lens form an image on a screen placed on the other side of the lens. The type of lens and its focal length …
- Where should the object be placed to get the virtual image in case of convex lens?…
- The lens used to correct long sightedness or hypermetropla.…
- Astigmatism for a human eye can be removed by using an additional:…