18. Find the position and magnification of the image of
an object placed at a distance of 8-0 cm in front of
a convex lens of focal length 10-0 cm. Is the image
erect or inverted ?
5.0 erect
Answers
Answer:
Image distance, (v) = -40 cm
Nature : Virtual and Erect
Magnification, (m) = +5
Step-by-step explanation:
Given,
Focal length of the lens, (f) = 10 cm
Object distance, (u) = - 8 cm
Let the distance of the image be 'v'. Then, we know that,
1/f = 1/v - 1/u
=> 1/v = 1/f + 1/u
=> 1/v = 1/10 + (-1) / 8
=> 1/v = 8/80 - 10/80
=> 1/v = -2/80
=> 1/v = -1/40
=> v = -40 cm
Since, the distance of image is negative hence it is on the side of object.
So the image is erect.
Nature : Virtual and Erect.
We know that,
Magnification, (m) = v/u
m = -40/(-8)
m = +5
Hence, the image is magnified to +5 times.
AnsWer ↓↓
Image distance, (v) = -40 cm
Nature : Virtual and Erect
Magnification, (m) = +5
Step-by-step explanation:
Given,
Focal length of the lens, (f) = 10 cm
Object distance, (u) = - 8 cm
Let the distance of the image be 'v'. Then, we know that,
1/f = 1/v - 1/u
=> 1/v = 1/f + 1/u
=> 1/v = 1/10 + (-1) / 8
=> 1/v = 8/80 - 10/80
=> 1/v = -2/80
=> 1/v = -1/40
=> v = -40 cm
Since, the distance of image is negative hence it is on the side of object.
So the image is erect.
Nature : Virtual and Erect.
We know that,
Magnification, (m) = v/u
m = -40/(-8)
m = +5
Hence, the image is magnified to +5 times.