Q. No.| 1. | infinity | 2. | \(+2~\text{D}\) |
| 3. | \(+20 ~\text{D}\) | 4. | \(+5~\text{D}\) |
1. infinite
2. zero
3. \(f/4\)
4. \(f/2\)
| 1. | \(2\) | 2. | \(1\) |
| 3. | No image will be formed | 4. | \(3\) |
A concave lens with a focal length of \(-25~\text{cm}\) is sandwiched between two convex lenses, each with a focal length of \(40~\text{cm}.\) The power (in diopters) of the combined lens system would be:
| 1. | \(55\) | 2. | \(9\) |
| 3. | \(1\) | 4. | \(0.01\) |
1. \(50\)
2. \(30\)
3. \(25\)
4. \(15\)
A point object is placed at a distance of \(60~\text{cm}\) from a convex lens of focal length \(30~\text{cm}\). If a plane mirror were put perpendicular to the principal axis of the lens and at a distance of \(40~\text{cm}\) from it, the final image would be formed at a distance of:
| 1. | \(30~\text{cm}\) from the plane mirror, it would be a virtual image. |
| 2. | \(20~\text{cm}\) from the plane mirror, it would be a virtual image. |
| 3. | \(20~\text{cm}\) from the lens, it would be a real image. |
| 4. | \(30~\text{cm}\) from the lens, it would be a real image. |
A plane-convex lens of unknown material and unknown focal length is given. With the help of a spherometer, we can measure the
| 1. | focal length of the lens. |
| 2. | radius of curvature of the curved surface. |
| 3. | aperture of the lens. |
| 4. | refractive index of the material. |
A biconvex lens has power \(P.\) It is cut into two symmetrical halves by a plane containing the principal axis. The power of one part will be:
| 1. | \(0\) | 2. | \(\dfrac{P}{2}\) |
| 3. | \(\dfrac{P}{4}\) | 4. | \(P\) |
Two identical glass \(\left(\mu_g = \frac{3}{2}\right )\) equiconvex lenses of focal length \(f\) each are kept in contact. The space between the two lenses is filled with water \(\left(\mu_w = \frac{4}{3}\right)\). The focal length of the combination is:
| 1. | \(\dfrac{f}{3}\) | 2. | \(f\) |
| 3. | \(\dfrac{4f}{3}\) | 4. | \(\dfrac{3f}{4}\) |
| 1. | \(10\) cm | 2. | \(20\) cm |
| 3. | \(40\) cm | 4. | zero |
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