An object of size 7.5cm is placed in front of a convex mirror of radius of curvature 25cm at a distance of 40cm. The size of the image should be
(1) 2.3 cm
(3) 1 cm
(2) 1.78 cm
(4) 0.8 cm
Answers
♣GIVEN♣
- f= 25/2 cm
- u= -40 cm
♣TO FIND♣
- The size of the image.
♣SOLUTION♣
➤ Mirror formula,
- 1/v +1/u =1/f
→ 1/v= 2/25 - (1/(-40))
→ 1v = 2/25+ 1/40 = 21/200
→ v = 200/21 cm
➤ Now,
- Magnification = h'/h = -(v/u)
→ h'/7.5= -(200/(21×(-40))
→ h' =25/17
→ h' =1.47 cm
So the image is virtual and diminished.
(2) 1.78cm
given :-
- size of object, h1 = +7.5cm
- radius of curvature, C = +25cm ( convex mirror)
- object distance, u = -40cm
we know that,
focal length,f = C/2
→ f = 25/2cm
_______________________
putting these values in the mirror formula:
1/f = 1/v + 1/u
→ 2/25 = 1/v - 1/40
→ 1/v = 2/25 + 1/40
→ 1/v = (16+5)/200
→ 1/v = 21/200
→ v = + 200/21 cm (in front of mirror)
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to find the size of image, we will calculate the magnification first,
now, magnification, m = -v/u
→ m = -(200/21)/-40
→ m = (200×1)/(21×40)
→ m = 5/21 cm
_______________________
we also have another formula for magnification, which is;
m = size of image,h2 / size of object, h1
→ 5/21 = h2 / 7.5
→ h2 = (5×7.5)/21
→ h2 = 37.5/21
→ h2 = 1.78 (approximately)
so, option c). 1.78 cm is correct
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✧ if the magnification has a plus sign, then the image is virtual and erect.
✧ if the magnification has a minus sign z then the image is real and inverted.
✧ if the magnitude of magnification is equal to 1, then the size of image is equal to size of object.
✧ if the magnitude of magnification is less than 1, then size of image is greater than size of object.
✧ if the magnitude of magnification is less than 1, then size of image is smaller than size of object