A concave mirror of focal length 15 cm can form a magnified, erect as well as inverted image of an object placed in front of it. Justify this statement stating the position of the object with respect to the pole of the mirror in both the cases for obtaining the images.

The image will be placed between pole P and focus F for magnified and erect image.

For magnified and inverted image the object is placed either at focus or in between F and C.

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Draw a ray diagram to show the path of the refracted ray in each of the following cases:

A ray of light incident on a concave lens is:

(i) passing through its optical centre.

(ii) parallel to its principal axis.

(iii) directed towards its principal focus.


For a concave lens,

i) The ray diagram for a ray of light passing through the optical centre of the concave lens will emerge without any deviation.


ii) For a ray of light directed towards the principal axis, after refraction from a concave lens, appears to diverge from the principal focus on the same side of the lens.

iii) For a ray of light directed towards its principal focus, goes parallel to its principal axis after refraction through the lens.

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To construct ray diagram we use two light rays which are so chosen that it is easy to know their directions after reflection from the mirror. List these two rays and state the path of these rays after reflection. Use these rays to locate the image of an object placed between centre of curvature and focus of a concave mirror.

In order to locate the image, any two rays can be chosen.

i) a ray which is parallel to the principal axis, after reflection will pass through the principal focus in case of a concave mirror and diverge from the principal axis in case of a convex mirror.

ii) A ray passing through the principal focus will emerge parallel to the principal axis after reflection.

When an object is placed between the centre of curvature and focus of a concave mirror, the ray diagram is as shown below:

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A 4 cm tall object is placed on the principal axis of a convex lens. The distance of the object from the optical centre of the lens is 12 cm and its sharp image is formed at a distance of 24 cm from it on a screen on the other side of the lens. If the object is now moved a little away from the lens, in which way (towards the lens or away from the lens) will he have to move the screen to get a sharp image of the object on it again? How will the magnification of the image be affected? 


Given that,

Object distance, u = −12 cm

Image distance, v = 24 cm

1 over straight f equals 1 over v minus 1 over u
1 over straight f space equals space 1 over 24 minus fraction numerator 1 over denominator open parentheses negative 12 close parentheses end fraction
1 over straight f space equals space 1 over 8
therefore space f space equals space 8 space c m

The focal length of the lens is 8 cm.

Now if the object is moved away from the lens, the screen has to be moved towards the lens. This is because when we move the object away from the lens, the object distance is increased. Hence, by the lens formula, the image distance decreases.

Magnification is given as,

straight m space equals space straight v over straight u
Since, the image distance (v) decreases, the value of magnification also decreases.

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A student wants to project the image of a candle flame on a screen 90 cm in front of a mirror by keeping the flame at a distance of 15 cm from its pole.

(a) Suggest the type of mirror he should use.

(b) Determine the linear magnification in this case.

(c) Find the distance between the object and its image.

(d) Draw ray diagram to show the image formation in this case.


a) Concave mirror should be used inorder to project the flame of the candle.

b) Linear magnification of the concave mirror is,



c) Distance between the object and the image = 90 - 15 = 75 cm

d) The image formation is as shown below:

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