Revision/ Ch7/ Q6

Revision/ Ch7/ Q6

  4:33 PM    Rohit

---

Q1

Assume a certain concave, spherical mirror has a focal length of 10.0 cm.

(a)   Locate the image and find the magnification for an object distance of 25.0 cm. Determine 

        whether the image is real or virtual, inverted or upright, and larger or smaller. Do the  

        same for object distances of

(b) 10.0 cm and

(c)  5.00 cm.

[16.7 cm, – 0.668, , – 10.0 cm, 2.00]

Q2
An object 3.00 cm high is placed 20.0 cm from a convex mirror with a focal length of magnitude 8.00 cm. Find

(a) the position of the image,

(b) the magnification of the mirror, and

(c) the height of the image.

[– 5 71 cm, 0.286, 0.858 cm]

Q3
When a woman stands with her face 40.0 cm from a cosmetic mirror, the upright image is twice as tall as her face. What is the focal length of the mirror?

[80.0 cm]

Q4
A coin 2.00 cm in diameter is embedded in a solid glass ball of radius 30.0 cm. The index of refraction of the ball is 1.50, and the coin is 20.0 cm from the surface. Find the position of the image of the coin and the height of the coin’s image.

[– 17.1 cm, 2.56 cm]

Q5
A converging lens of focal length 10.0 cm forms images of an object situated at various distances.
(a)   If the object is placed 30.0 cm from the lens, locate the image, state whether it’s real or 
       virtual, and find its magnification.
(b) Repeat the problem when the object is at 10.0 cm and
(c) again when the object is 5.00 cm from the lens.

[+ 15.0 cm, – 0.500, , – 10.0 cm, + 2.00]

Q6/ [Use lens-maker’s equation]

A thin diverging glass (index = 1.50) lens with R1 = - 3.00 m and R2 = - 6.00 m is surrounded by air. An arrow is placed 10.0 m to the left of the lens.

(a) Determine the position of the image. Repeat part (a) with the arrow and lens immersed in

(b) water (index = 1.33) and

(c) a medium with an index of refraction of 2.00.

(d) How can a lens that is diverging in air be changed into a converging lens?

[5.45 m to the left of the lens, 8.24 m to the left of the lens, 17.1 m to the left of the lens, by surrounding the lens with a medium having a refractive index greater than that of the lens material]