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प्रश्न
Complete the following diagrams shown in the below figure by drawing the reflected ray for each incident ray.
उत्तर
A light ray when produced backwards passes through principal focus as shown in the problem figure. We draw the normal through the centre of curvature at the point of incidence and draw the reflected ray at an angle equal to the angle of incidence thus following the laws of reflection. The reflected ray is parallel to the principal axis. The other ray is passing through the centre of curvature as shown in the problem figure. This ray retraces its incident path because it strikes the mirror normally i.e. 90 degrees. These two reflected rays when produced backwards coincide at a point where the image is formed. The image, A'B' is virtual, erect, and diminished in size.
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संबंधित प्रश्न
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Match the items given in Column I with one or more items of Column II.
| Column I | Column II | ||
| (a) | A plane mirror | (i) | Used as a magnifying glass. |
| (b) | A convex mirror | (ii) | Can form image of objects spread over a large area. |
| (c) | A convex lens | (iii) | Used by dentists to see enlarged image of teeth. |
| (d) | A concave mirror | (iv) | The image is always inverted and magnified. |
| (e) | A concave lens | (v) | The image is erect and of the same size as the object. |
| (vi) | The image is erect and smaller in size than the object. |
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- the position of the image
- the focal length of the convex mirror.
Define the term Centre of curvature.
