An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm find the position and nature of the image
Answers
Question
An object is placed at a distance of 10cm from a convex mirror of focal length 15cm. Find the position and nature of the image ?
ANSWER
Given :-
An object is placed at a distance of 10cm from a convex mirror of focal length 15cm
Required to find :-
Position and nature of the image ?
Solution :-
Object distance (u) = -10cm
This is due to the rule of sign convention ! as all distance are measured with respective the pole.
Focal of a convex mirror is always positive !
So,
Focal length (f) = 15cm
Now,
We know that;
(1)/(f) = (1)/(u)+(1)/(v)
(1)/(15) = (1)/(-10)+(1)/(v)
(1)/(v) = (1)/(15) - (1)/(-10)
(1)/(v) = (1)/(15)+(1)/(10)
(1)/(v) = (2+3)/(30)
(1)/(v) = (5)/(30)
v = (30)/(5)
v = 6 cm
Hence,
Image distance (v) = 6 cm
By drawing the ray diagram for the above situation using the information formulated from the above calculations. We can say that;
The position of the image between Pole and Focus.
The nature of image is Erect because it forms an upright image, Virtual because image is formed by diverging of the light rays, Image is considered to be as diminished as the size of the image is relatively smaller than the height of the object.
Answer:
It is to be remembered that a convex mirror always forms the virtual, erect and diminished image.
Given: Object distance, u=−10cm Focal length, f=15cm
To find: Image distance, v
From mirror formula :
v
1
+
u
1
=
f
1
v
1
−
10
1
=
15
1
v
1
=
15
1
+
10
1
=
150
25
v=6cm
Hence image will be formed 6 cm beyond the mirror, i.e virtual magnification
m=
u
−v
=
−10
−6
=0.6
Hence it will be erect and diminished.
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