A concave mirror is used for images formation for different positions of an object. what inferences can be drawn about the following when an object is placed at a distance of 10cm from the pole of a concave mirror of focal length 15cm.
1) position of the image
2)size of the image
3) nature of the image
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Given
To Find
Solution
● Mirror Formula :- 1/v + 1/u = 1/f
● Size of the image :- M = hᵢ/hₒ= -v/u
✭ Image Distance :
→ 1/v + 1/u = 1/f
→ 1/v + 1/-10 = 1/-15
→ 1/v - 1/10 = - 1/15
→ 1/v = 1/10 - 1/15
→ 1/v = 3/30 - 2/30
→ 1/v = (3-2)/30
→ 1/v = 1/30
→ v = 30 cm
✭ Image Size :
→ -v/u = hᵢ/hₒ
→ -30/-10 = hᵢ/hₒ
→ hᵢ = 3 cm
∴ Thr image is virtual, erect, placed at 30 cm, magnified 3 times with the image size 3 cm
Given:-
•A concave mirror is used for images formation for different positions of an object..
•A object is placed at a distance of 10 cm from the pole of a concave mirror of focal length 15 cm..
To Prove:-
•Position of the image
•Size of the image
•Nature of the image
Solution:-
•Focal length = - 15 cm
•Object Distance = - 10 cm
•To Find Image distance,
•Using formula,
[tex] \leadsto{ \boxed{ \boxed{ \boxed{ \bold{ \red{ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} }}}}}}[/tex]
[tex] \leadsto{ \boxed{ \boxed{ \boxed{ \bold{ \frac{1}{v} = \frac{1}{10} - \frac{1}{15} }}}}}[/tex]
[tex] \leadsto{ \boxed{ \boxed{ \boxed{ \bold{ \frac{1}{v} = \frac{3}{30} - \frac{2}{30} }}}}}[/tex]
[tex] \leadsto{ \boxed{ \boxed{ \boxed{ \bold{ \frac{1}{v} = \frac{1}{30} }}}}}[/tex]
[tex] \leadsto{ \boxed{ \boxed{ \boxed{ \bold{ \red{ v = 30 \: cm}}}}}}[/tex]
•To Find Size of image,
•h1 = height of the image
•h2 = height of the object
•Using Formula,
[tex] \leadsto{ \boxed{ \boxed{ \boxed{ \bold{ \red{ \frac{ - v}{u} = \frac{h_
1}{h_2}}}}}}}[/tex]
[tex] \leadsto{ \boxed{ \boxed{ \boxed{ \bold{ \frac{ - 30}{10} = \frac{h_
1}{h_2}}}}}}[/tex]
[tex] \leadsto{ \boxed{ \boxed{ \boxed{ \bold{ \red{h_1 = 3 \: cm }}}}}}[/tex]
•The image is virtual and erect
•The image distance is 30 cm
•The size of the image is 3 cm