Question

1. which of the following characteristics of the image formed by a diverging lens are correct?

a. virtual and diminished
b. real and diminished
c. virtual and magnified
d. real and magnified

1. light from an illuminated object passes through a converging lens and is reflected along its original path to form an image at the same position as the object by a plane mirror placed just behind the lens. if the distance between the object and the plane mirror is 20 cm, determine the focal length of the lens.

a. 40 cm
b. 20 cm
c. 10 cm
d. 5 cm

1. the advantage of the galilean telescope over the astronomical telescope is that in the galilean telescope, the

a. final image is sharper.
b. angular magnification is greater.
c. field of view is wider.
d. final image is erect.

1. a singer emits a note of frequency 200 hz. calculate the wavelength of the note. speed of sound in air = 340 m s⁻¹

a. 17.0 m
b. 6.0 m
c. 1.7 m
d. 0.6 m

1. echo is a phenomenon that shows that sound waves

a. are mechanical.
b. are longitudinal.
c. can be polarized.
d. can be reflected.

1. which of the following statements about force fields is correct?

a. electrostatic, gravitational and magnetic forces are always attractive.
b. electric, gravitational and magnetic fields obey inverse square laws.
c. field lines are real but their corresponding fields are imaginary.
d. field lines and their corresponding fields are both real.

1. which of the following statements about a satellite is/are correct? a satellite

i. describes a circle about the earth if its escape velocity is 8 km s⁻¹.
ii. cannot escape from the earth’s surface with a velocity less than 8 km s⁻¹.
iii. describes an ellipse about the earth if its escape velocity is greater than 11 km s⁻¹.
a. i only
b. ii only
c. i and ii only
d. ii and iii only

Explanation:

Question 30

A diverging lens always forms a virtual, upright, and diminished image regardless of the object's position (within the rules of lens imaging). So the correct characteristics are virtual and diminished.

Step1: Analyze the setup

The light passes through the converging lens, reflects off the plane mirror, and retraces its path. This means the light rays from the object, after passing through the lens, are incident normally on the plane mirror (so they retrace). For a converging lens, when the object is at 2F (twice the focal length), the image is also at 2F on the other side. But here, the object and its image (formed by the lens - mirror system) are at the same position, so the distance from the object to the lens is equal to the focal length? Wait, no. Wait, the plane mirror is just behind the lens. The light goes through the lens, hits the mirror, and comes back. So the image formed by the lens (before reflection) should be at the mirror's position (since after reflection, it comes back to the object). Wait, the distance between object and mirror is 20 cm. So the object is at a distance \( u \) from the lens, and the image formed by the lens (without mirror) would be at \( v \), but with the mirror, the light reflects back. So for the light to retrace, the image formed by the lens must be at the mirror's position (so that the rays hit the mirror normally). So the object distance \( u \) and image distance \( v \): since the mirror is behind the lens, the distance from object to mirror is 20 cm, so the distance from object to lens is \( u \), and from lens to mirror is \( 20 - u \). But for the lens, when the light retraces, the image formed by the lens is at the mirror, so \( v = 20 - u \)? Wait, no. Wait, the key is that when the light passes through the lens, forms an image, then reflects off the mirror (which is at the image position) and comes back. So the object distance for the lens is equal to the image distance? Wait, no. Wait, the plane mirror is placed just behind the lens. So the distance from the object to the mirror is 20 cm, so the distance from the object to the lens is \( d \), and from the lens to the mirror is \( 20 - d \). But for the light to retrace, the image formed by the lens must be at the mirror (so that the rays are incident normally on the mirror). So for the lens, the object is at \( u = d \), and the image is at \( v = 20 - d \). But when the light reflects back, it's like the image becomes the object for the return journey. But for the light to retrace, the lens must form an image at the mirror, so that the reflected rays (which are now going back) will pass through the lens and form an image at the original object position. Wait, maybe a simpler approach: when the light passes through the lens, then reflects off the mirror (which is at the focal point? No). Wait, the condition for the light to retrace its path is that the rays are incident normally on the mirror, which means the image formed by the lens is at the mirror's position. So the object and its image (formed by the lens) are such that the mirror is at the image position. But since the mirror is just behind the lens, the distance from the object to the mirror is 20 cm, so the distance from the object to the lens is \( u \), and the distance from the lens to the mirror is \( 20 - u \). But for the lens, when the image is at the mirror, \( v = 20 - u \). But also, for the light to retrace, the lens must form an image at the mirror, so that the reflected rays (which are parallel? No, wait, if the object is at the focal point of the lens, the rays after the lens are parallel. Then, a plane mirror would reflect them back parallel, and they would retrace. Wait, no: if the object is at the focal point of the converging lens, the rays from the obj…

A Galilean telescope uses a converging objective lens and a diverging eyepiece, while an astronomical telescope uses a converging objective and a converging eyepiece. The main advantage of the Galilean telescope is that the final image is erect (since the diverging eyepiece inverts the image again, canceling the inversion from the objective), whereas the astronomical telescope produces an inverted image. The other options: sharper image (not necessarily), greater angular magnification (astronomical can have greater), wider field of view (astronomical has wider). So the correct advantage is the final image is erect.

Answer:

A. Virtual and diminished

Question 31