Question

11 What type of image is formed when an object is placed at a distance of 1.5 focal lengths from a convex mirror?
a. erect and virtual
b. inverted and virtual
c. erect and real
d. inverted and real

12 Where is the image located when an object is 60 cm from a convex mirror with a focal length of20 cm?
a. 15 cm behind
b. 30 cm behind
c. 60 cm behind
d. 15 cm in front

13. If the sun is 150 million km away from the earth, how long does it take sunlight to reach the earth
a. 0.5 s
b. 1500 s
c. 45 s
d. 500 s
14 The critical angle for total internal reflection at an air-water interface is approximately 48 degrees. In which of the following situations will total internal reflection occur?
a. light incident in water at 40 dgrees
b. light incident in water at 55 dgrees
c. light incident in air at 40 dgrees
d. light incident in air at 55 dgrees
15 How many diopters are there for for a converging lens with a focal length of 0.4m?

a. −2.5
b. −0.4
c. +0.4
d. +2.5

Answer 1

Answer on Question 51680, Physics, Optics

11. What type of image is formed when an object is placed at a distance of 1.5 focal lengths from a convex mirror?

a) erect and virtual

b) inverted and virtual

c) erect and real

d) inverted and real

Answer:

The image on a convex mirror is always virtual, and as we can see when an object is placed at a distance of 1.5 focal lengths from a convex mirror, the image is virtual and erect. Correct answer: a) erect and virtual.

12. Where is the image located when an object is $60cm$ from a convex mirror with a focal length of $20cm$ ?

a) $15cm$ behind

b) $30cm$ behind

c) $60cm$ behind

d) $15cm$ in front

Solution:

From the mirror equation we have:

$$\begin{array}{l} \frac{1}{d_{image}} + \frac{1}{d_{object}} = \frac{1}{-f}, \\ \frac{1}{d_{image}} + \frac{1}{60cm} = \frac{1}{-20cm}, \\ \frac{1}{d_{image}} = -\frac{1}{20cm} - \frac{1}{60cm} = -\frac{1}{15cm}, \\ d_{image} = -15cm. \end{array}$$

The negative sign of $d_{image}$ indicate that image is located behind the convex mirror.

Answer: a) $15cm$ behind.

13. If the Sun is 150 million kilometers away from the Earth, how long does it take sunlight to reach the Earth?

a) $0.5s$

b) $1500s$

c) $45s$

d) $500s$

**Solution:**

In order to find time that needs sunlight to reach the Earth we divide the distance to the Sun by the speed of light and obtain:

$$t = \frac{d}{c} = \frac{1.5 \cdot 10^{11}m}{3 \cdot 10^{8} \frac{m}{s}} = 500s.$$

Answer: d) $500s$

14. The critical angle for total internal reflection at an air-water interface is approximately $48{}^{\circ}$. In which of the following situations will total internal reflection occur?

a) light incident in water at $40{}^{\circ}$

b) light incident in water at $55{}^{\circ}$

c) light incident in air at $40{}^{\circ}$

d) light incident in air at $55{}^{\circ}$

Solution:

The total internal reflection occurs when light attempts to move from a medium having a given refractive index to a medium having a lover refractive index (in our case from water with $n_1 = 1.33$ to air with $n_2 = 1.0$ ).

As we can see in the picture for $\theta_{1} > \theta_{c}$ there is no reflected ray. Thus, in order to occur the total internal reflection we need b) light incident in water at $55{}^{\circ}$ .

Answer: b) light incident in water at $55{}^{\circ}$ .

15. How many diopters are there for a converging lens with a focal length of $0.4m$ ?

a) $-2.5$

b) $-0.4$

c) $+0.4$

d) $+2.5$

Solution:

The power of a lens is defined as the reciprocal of its focal length in meters:

$$P = \frac {1}{f} = \frac {1}{0.4m} = +2.5diopters$$

Because we have a converging lens (a positive lens) the correct answer is d) +2.5

Answer: d) +2.5

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