Text Solution Solution : As the object is moved away from a convex mirror, the distance of virtual image from the mirror (formed behind it between pole and focus) increases i.e., the image shifts from the pole towards the focus and the size of the image gradually decreases. When the object is at infinity (very far), the image is at its focus. The image is always erect and virtual.
There are only two possibilities of position of object in the case of a convex mirror, i.e. object at infinity and object between infinity and pole of a convex mirror. Object at infinity: When the object is at the infinity, a point sized image is formed at principal focus behind the convex mirror. Fig: Object at infinity Properties of image: Image is highly diminished, virtual and erect. Object between infinity and pole: When the object is between infinity and pole of a convex mirror, a diminished, virtual and erect image is formed between pole and focus behind the mirror. Fig: Object between infinity and P Properties of image: Image is diminished, virtual and erect.
Uses of Concave Mirror:
Uses of Convex Mirror:
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An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Converging lens means a convex lens. As the distances given in the question are large, so we choose a scale of 1: 5, i.e., 1 cm represents 5 cm. Therefore, on this scale 5 cm high object, object distance of 25 cm and focal length of 10 cm can be represented by 1 cm high, 5 cm and 2 cm lines respectively. Now, we draw the ray diagram as follows:(i) Draw a horizontal line to represent the principal axis of the convex lens.(ii) Centre line is shown by DE.(iii) Mark two foci F and F' on two sides of the lens, each at a distance of 2 cm from the lens.(iv) Draw an arrow AB of height 1 cm on the left side of lens at a distance of 5 cm from the lens.(v) Draw a line AD parallel to principal axis and then, allow it to pass straight through the focus (F') on the right side of the lens.(vi) Draw a line from A to C (centre of the lens), which goes straight without deviation.(vii) Let the two lines starting from A meet at A'.(viii) Draw AB', perpendicular to the principal axis from A'.(ix) Now AB', represents the real, but inverted image of the object AB.(x) Then, measure CB' and A'B'. It is found that CB' = 3.3 cm and A'B' = 0.7 cm. (xi) Thus the final position, nature and size of the image A'B' are: (a) Position of image A'B' = 3.3 cm × 5 = 16.5 cm from the lens on opposite side. (b) Nature of image A’B’: Real and inverted. (c) Height of image A'B': 0.7 × 5 = 3.5 cm, i.e., image is smaller than the object. |