An object is placed at a distance of 50cm infront of a converging mirror of focal length 20cm. Find the position,nature and size of image?
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A converging mirror is also known as a 'concave' mirror.
Now, using the sign convention for spherical mirrors, we can say that u (object distance) and f(focal length) are negative.
Or,
u = -50cm
f = -20cm
Now, using the mirror formula,
1/f = 1/u + 1/v
-1/20 = -1/50 + 1/v
-1/20 + 1/50 = 1/v
-5 + 2/ 100 = 1/v ......(taking the LCM of 20,50 as 100)
-3/100 = 1/v
-100/3 = v
-33.33 cm =v
Now, we know magnification produced by spherical mirror is
m= -v/u
m= - (- 100/3 * -1/50) .... ( I took v= - 100/3 instead of -33.33 for ease of calculation)
m= -2/3=- 0.66
Therefore,
Position of the image: 33.33 cm from the mirror. The image is formed between C & F.
Nature of the image : The image is real, inverted and diminished.
Size of the image : The image is diminished by a factor of 0.66.
PS- I tried to make it as simple as I could. If you have any doubts, feel free to ask. Do check my calculations! ;P
NISHTHA.
Now, using the sign convention for spherical mirrors, we can say that u (object distance) and f(focal length) are negative.
Or,
u = -50cm
f = -20cm
Now, using the mirror formula,
1/f = 1/u + 1/v
-1/20 = -1/50 + 1/v
-1/20 + 1/50 = 1/v
-5 + 2/ 100 = 1/v ......(taking the LCM of 20,50 as 100)
-3/100 = 1/v
-100/3 = v
-33.33 cm =v
Now, we know magnification produced by spherical mirror is
m= -v/u
m= - (- 100/3 * -1/50) .... ( I took v= - 100/3 instead of -33.33 for ease of calculation)
m= -2/3=- 0.66
Therefore,
Position of the image: 33.33 cm from the mirror. The image is formed between C & F.
Nature of the image : The image is real, inverted and diminished.
Size of the image : The image is diminished by a factor of 0.66.
PS- I tried to make it as simple as I could. If you have any doubts, feel free to ask. Do check my calculations! ;P
NISHTHA.
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