An object 4cm high is placed at a distance of 16cm from a concave mirror which produces a real image 5cm in high. Find the position of the image. What is the nature of the image formed.
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15
m = - v / u = h (image) / h (object)
-v / -16 = -5 / 4
v = -20 cm
m = -5 / 4, So the image is real, inverted because m is negative and it is magnified because m > 1
Answered by
1
Answer:
FOCAL LENGTH = -7.74 & IMAGE DISTANCE = -15cm
Explanation:
Solution:
Height of object h₁ = 4cm
Height of image h₂ = 5cm
★ Object distance, u = = 16cm
★ Image distance = v
Magnification h₁ h₂ = =
5 4 -v -16
• −16 × 5 = −4 × −v
-60 = 4v
-60
4
V= 15cm
★ Using mirror formula :
1/u + 1/u = 1/f
1/- 16 + 1/- 15 = 1/f
((15) + (16))/- 240 = 1/f
31/- 240 = 1/f
f = - 240/31
f = -7.74cm
Hence,
FOCAL LENGTH = -7.74cm & IMAGE DISTANCE = -15cm
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