An object,4.0cm in size is placed at 25cm in front of a concave mirror of focal length 15.0cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image?
Answers
Answer: Size of image is 6 cm
Explanation:
focal length = f = - 15 cm
object distance = u = -25 cm
image distance = v = ?
hi = ?
ho = 4 cm
Mirror formula,
1/v + 1/u = 1/f
1 over f space equals space 1 over u plus 1 over v
rightwards double arrow fraction numerator 1 over denominator negative 15 end fraction equals fraction numerator 1 over denominator negative 25 end fraction plus 1 over v
rightwards double arrow space fraction numerator 1 over denominator negative 15 end fraction plus 1 over 25 space equals space 1 over v
rightwards double arrow space 1 over v equals fraction numerator 25 minus 15 over denominator negative 375 end fraction
rightwards double arrow space v space equals negative space 37.5 space c m space
m equals fraction numerator negative v over denominator u end fraction equals h subscript i over h subscript o
rightwards double arrow space m space equals fraction numerator negative left parenthesis negative 37.5 right parenthesis over denominator negative 25 end fraction equals h subscript i over 4
h subscript i space equals space 150 over 25 equals space minus 6 space c m
Thus, to get image of object sharp it has to be placed at 37.5 cm
Size of image is 6 cm
As the height of object is more and negative, nature of image is real, inverted and magnified.
Answer:
focal lengt(f)= -15
object distance (u) = -25
image distance= (v)=?
h1=?
ho=4 cm
mirror formula,
=1/v+ 1/u=1/f
=1/f = 1/u+1/v
=1/15=1/-25 + 1/v
=1/v=25-15/-375
v= -37.5cm
m= -(-37.5)/-25 =h1/4
h1= 150/25 = -6cm
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