A point object is placed at a distance of 30 cm from a convex mirror of focal length 30 cm. The image will form at
1. infinity
2. pole
3. focus
4. 15 cm behind the mirror
The figure shows two rays A and B being reflected by a mirror and going as A' and B'. The mirror,
1. is plane
2. is convex
3. is concave
4. may be any spherical mirror
The image formed by a concave mirror:
1. | is always real |
2. | is always virtual |
3. | is certainly real if the object is virtual |
4. | is certainly virtual if the object is real |
The figure shows three transparent media of refractive indices \(\mu_1,~\mu_2\) and \(\mu_3\). A point object \(O\) is placed in the medium \(\mu_2\). If the entire medium on the right of the spherical surface has refractive index \(\mu_1\), the image forms at \(O'.\) If this entire medium has refractive index \(\mu_3\), the image forms at \(O''.\) In the situation shown,
1. | \(O'\) and \(O''.\) | the image forms between
2. | \(O'.\) | the image forms to the left of
3. | \(O''.\) | the image forms to the right of
4. | \(O'\) and the other at \(O''.\) | two images form, one at
Four modifications are suggested in the lens formula to include the effect of the thickness t of the lens. Which one is likely to be correct?
1. \(\frac{1}{v}-\frac{1}{u}=\frac{t}{u f}\)
2. \(\frac{t}{v^{2}}-\frac{1}{u}=\frac{1}{f}\)
3. \(\frac{1}{v-t}-\frac{1}{u+t}=\frac{1}{f}\)
4. \(\frac{1}{v}-\frac{1}{u}+\frac{t}{u v}=\frac{t}{f}\)
A double convex lens has two surfaces of equal radii \(R\) and refractive index \(m=1.5.\) We have:
1. \(f=R/2\)
2. \(f=R \)
3. \(f=-R\)
4. \(f=2R\)
A point source of light is placed at a distance of \(2 f\) from a converging lens of focal length \(f.\) The intensity on the other side of the lens is maximum at a distance:
1. | \(f\) | 2. | \(f\) and \(2 f\) | between
3. | \(2 f\) | 4. | \(2 f\) | more than
A parallel beam of light is incident on a converging lens parallel to its principal axis. As one moves away from the lens on the other side on its principal axis, the intensity of light:
1. | remains constant |
2. | continuously increases |
3. | continuously decreases |
4. | first increases then decreases |
A symmetric double convex lens is cut in two equal parts by a plane perpendicular to the principal axis. If the power of the original lens was \(4 D,\) the power of a cut-lens would be:
1. \(2 D\)
2. \(3 D\)
3. \(4 D\)
4. \(5D\)
A symmetric double convex lens is cut in two equal parts by a plane containing the principal axis. If the power of the original lens was \(4~\text{D},\) the power of a divided lens will be:
1. \(2~\text{D}\)
2. \(3~\text{D}\)
3. \(4~\text{D}\)
4. \(5~\text{D}\)