Use the mirror equation to deduce that: (a) an object placed between f and 2 f of a concave mirror produces a real image beyond 2 f. (b) a convex mirror always produces a virtual image independent of the location of the object. (c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole. (d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.[Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams]
Answers
a) Mirror Formula : 1/v+1/u=1/f
For concave mirror:
f<0, as object always placed on the left side of mirror so u<0
According to question:
f<u<2f [ object lies between f and 2f]
1/2f>1/u>1/f
or
-1/2f<-1/u< -1/f
add 1/f on both sides :
1/f-1/2f< 1/f-1/u<0------------(1)
From lens formula : 1/f=1/v-/1u
we get : 1/2f <1/v
or v>2f
as f is negative , 2f is also negative and hence v will be also negative.
Nature of image : Real and lies between beyond 2f
b)For convex mirror :
f is positive
f>0 and u<0
from lens formula : 1/v =1/f-1/u
as f>0 and u<0
value 1/v >0 or v>0 or v is always positive.
The image formed is virtual . It does not depend on location of object.
c) For convex mirror f>0, u<0
from lens formula : 1/v=1/f-1/u
1/v>1/f or v<f
Thus image is always located between pole and focus of mirror.as v<|u|so image is always diminished in size.
d) For concave mirror f<0
As object is placed between pole and focus.
f<u<0
1/f-1/u>0
From lens formula : 1/f-1/u=1/v
v>0
That means v is positive and image is formed on right virtual .
1/v<1/u
v>|u| so image is enlarged.
Answer:
Explanation:a) Mirror Formula : 1/v+1/u=1/f
For concave mirror:
f<0, as object always placed on the left side of mirror so u<0
According to question:
f<u<2f [ object lies between f and 2f]
1/2f>1/u>1/f
or
-1/2f<-1/u< -1/f
add 1/f on both sides :
1/f-1/2f< 1/f-1/u<0------------(1)
From lens formula : 1/f=1/v-/1u
we get : 1/2f <1/v
or v>2f
as f is negative , 2f is also negative and hence v will be also negative.
Nature of image : Real and lies between beyond 2f
b)For convex mirror :
f is positive
f>0 and u<0
from lens formula : 1/v =1/f-1/u
as f>0 and u<0
value 1/v >0 or v>0 or v is always positive.
The image formed is virtual . It does not depend on location of object.
c) For convex mirror f>0, u<0
from lens formula : 1/v=1/f-1/u
1/v>1/f or v<f
Thus image is always located between pole and focus of mirror.as v<|u|so image is always diminished in size.
d) For concave mirror f<0
As object is placed between pole and focus.
f<u<0
1/f-1/u>0
From lens formula : 1/f-1/u=1/v
v>0
That means v is positive and image is formed on right virtual .
1/v<1/u
v>|u| so image is enlarged.
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