Question

3. An object is placed at the following distances from a concave mirror of focal length 15 cm.
(a) 10 cm
(b) 20 cm
(c) 30 cm
(d) 40 cm
Which position of the object will produce:
(i) Virtual image
(ii) A diminished real image
(i) An enlarged real image
(i) An image of same size.​

Answer 1

Answer:

Explanation:

radius of curvature, R= 2f  ; f=focal length

∴ here R=2x15=30cm  : since f=15cm

magnification,m=-v/u  ; {v-image distance, u-object distance}

m >1 ⇒ the image is magnified

m=1 ⇒size of image =size of object

m<1⇒image is diminished

[C- centre of curvature, F- focus, P-pole]

(a) object at 10cm ie, between F and P----image is formed behind the mirror and it will be virtual and erect

(b)object at 20cm i.e, between C and F----image is formed beyond C and it will be real and inverted (magnified image)

(c) object at 30 cm i.e, at C----image is formed at C and it will be real and inverted(image of same size)

(d)object at 40cm i.e, beyond C -----image is formed between C and F and it will be real  and inverted(diminished image)

Answer 2

Explanation:

given,

focal length, f= 15 cm= 0.15 m

here if the obj is placed between the pole and focal length of the concave mirror, the following characteristics will be observed,,

• virtual image
• enlarged
• erect

if it is placed between focal point and the centre of curvature,

we will see the following characteristics,

• real image
• diminished
• inverted

if we place the obj in the centre of curvature,

• real
• equal size
• inverted

if we place the object between centre of curvature and infinity,

we will see the following characteristics,

• real image
• enlarged
• inverted

now,

For 10 cm,

object is between focal point and pole

so,

• virtual
• enlarged
• erect

for 20 cm,

object is between focal point and centre of curvature

• real
• diminished
• inverted

For 30 cm,

object is in the centre of curvature

• real
• equal size
• inverted

for 40 cm,

object is between centre of curvature and infinity,

• real
• enlarged
• inverted