Question

An object is imaged on to a photographic sensor using a thin positive lens. The distance between the object and its image is 18 cm, such that the image size is half the size of the object.

(a) Determine the distance between the lens and the image on the photographic sensor.

(b) A glass plate having plane parallel faces, is 0.5 cm thick and has a refractive index of 1.5. This is inserted between the lens and the photographic sensor, such that the distance between the 1st planar surface and the thin lens is 1.5 cm. Determine the new position of the image. {Hint: Consider the glass plate as a thick lens with surfaces having the appropriate radii of curvature}

Answer 1

(a)The magnification of the image is 1 m=_=- u=2v The distance between the image and the object is, uV 18 cm 2v 18 cm 3v- 18 am V-6 cm 2v -2(6 cm) u=12 cm Hence, the distance between the lens and image is 6 cm (b)The diagram of the ray is as follows r P R. i From the diagram, ANPO NP sini OP From the diagram, ANPI NP sinr IP

From the Snell's law, for air glass medium n sinin sinr NP NP (1) OP IP OP ON IP IN ON IN The normal shift is ON ON OI ON IN =ON- = So, the normal shift of the image is OI (0.5 cm 1-- =0.17 cm 1.5 position of the image is So, the new v6 cm OI 6 cm+0.17 am-6.17 cm position of the image is 6.17 cm So, the new