(а) Draw a ray diagram to show the formation of image of an object placed between infinity and the optical centre of a concave lens.
(b) A concave lens of focal length 15 cm forms an image 10 cm from the lens. Calculate
(i) the distance of the object from the lens.
(ii) the magnification for the image formed.
(iii) the nature of the image formed.
(a) Image formation when the object is at any position between infinity and optical centre:
Position of image: Between F and optical centre
Nature of image: Virtual, erect and diminished.
(b) f = -15 cm v = -10 cm
(i) Using, 1/f = 1/v - 1/u, we get
1/u = 1/v - 1/f = 1/-10 - 1/-15 = - 1/30
∴ u = -30 cm
Therefore the object is at 30 cm from the concave lens on its left side.
(ii) m = -10/-30 = 1/3
(iii) m is +ve, so image is erect, v is -ve, so image is virtual. As m < 1, image is diminished.