Question

an object of size 7 centimetre is placed at 27 CM in front of a concave mirror of focal length 18cm. what distance from the the mirror should a screen be placed so that a sharp focussed image can be of time find the size and nature of the image. please help me​

Answer 1

${\tt{\green{\underline{\underline{\huge{Question:-}}}}}}$

• An object of size 7 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focused image can be obtained ? Find the size and the nature of the image.

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${\tt{\red{\underline{\underline{\huge{Answer:}}}}}}$

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$\mathrm{\boxed{\boxed{\pink{→ Mirror \: Formula ✔}}}}$

• $\frac{1}{v} + \frac{1}{u} = \frac{1}{f} .....(1)$

$\mathrm{\boxed{\boxed{\red{→ So, <strong><u>\</u></strong><strong><u>:</u></strong><strong><u>u </u></strong>= \:-27cm✔}}}}$

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$\sf\color{blue}{→Concave \:mirror\: have\: negative\: focal\: length}$

$\mathrm{\boxed{\boxed{\red{→ So, \:f= \:-18cm}}}}$

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Putting the values

$\frac{1}{v} + \frac{1}{ - 27} = \frac{1}{ - 18}$

$\sf\color{blue}{We \: get \: <strong>v </strong>\: = -54cm}$

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$\mathrm{\boxed{\boxed{\pink{→ Magnification \: Formula ✔}}}}$

• $\frac{h2}{h1} = \frac{ - v}{u}$
• $\frac{h2}{7.0} = - \frac{ - 54}{ - 27}$

$\mathrm{\boxed{\boxed{\red{→ So, \:h2= \:-14.0cm}}}}$

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Image will be real and inverted and will be of size 14 cm.

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${\huge{\underline{\small{\mathbb{\blue{HOPE\:HELP\:U\:BUDDY :)}}}}}}$

Answer 2

Answer:

We know that, for a concave mirror the focal length will be negative.

Therefore, given: u = -27cm     f = -18cm    h(object size) = 7cm

( Note that the distance at which you will obtain sharp image will be your "v" i.e.  the object distance for that  image formation which is required in the question)

hence,

1/v + 1/u = 1/f                                       (mirror formula)

1/v + (1/(-27)) = 1/(-18)

1/v  =   1/27 + 1/-18                              ( LCM of 18 and 27 is 54)

1/v  =  1/(-54)

v     =  -54cm

image distance = -54cm

hence, image is formed on the same side.

now we know,

h'/h = -v/u                                             (h' denotes object height)

h'/7 = -(-54)/-(27)

h'    =  7 * (-2)

h'    =  -14cm

hence, image is inverted (since h' is negative) and it is on the same side of that of the object (since v is negative).

HOPE IT HELPS .