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VISN1111 (2013)

VISN1111 Geometrical and Physical Optics 2013

Tutorial-7

1. An object of height 25mm is located 250mm from the eye. Model the eye as a

single refracting surface of power 60D separating air from the aqueous medium

of refractive index 4/3, and calculate the image location and height.

The same object is now viewed through a +10D thin lens located 100mmfrom the surface of the eye represented by the model above. Calculate the

object location for the +10D lens if the image formed by this lens is 250mm

from the surface of the reduced eye. What is the image height? Use this result to

find the final image height on the back of the eye.

Calculate the magnifying power of the +10D lens and show that it is simply the

ratio of the two final image heights calculated in the previous part of thequestion.

2. An object 5 mm long is situated 10 cm from a lens of power +7.50 D. If an

observer views the image through the lens and the distance between the lens and

his eye is 8 cm, determine the visual angle subtended by the image.

Ans. 2.39

3. What is the angular magnification produced by a +12.00 D lens 4 cm in front of

an emmetropic eye which is a) unaccommodated b) accommodated 5 D

Ans. 3x, 3.65x

4. A patient can barely read 5 M print at 40 cm distance. He likes to read 2 M prints.

Find the EVD and EVP1. Ans: 16 cm & 6.25 D

5. A patient can barely see 6 M print at 40 cm distance. She is prescribed a 10 D

magnifier which she holds at a 10 cm distance from the page. What size print can

she see clearly? Ans: 1.5 M

6. A CCTV (Closed Circuit Television) may be used as a low vision aid! Patients can

read at a comfortable distance. Contrast may be changed to white letters on a black

background! Your patient can read 6 M print at 30 cm distance. She would like to

read 1 M print at 40 cm distance. By how much should the CCTV magnify the

characters? Ans: 8