Question

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An object 5 cm in size is placed at 12 CM in front of a concave mirror of radius of curvature of 20 CM at what distance from the mirror should a screen be placed in order to obtain a sharp image? find the nature and size of the L.

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Answer 1

Answer:

It is given that :-

Height of object = 5cm

Object distance = u = 12cm

Focal length of mirror = R /2 = 20/2 = 10cm.

Image distance = v = ?

Height of image = ?

Now, according to mirror formula :

After inputting the known values we get :

✠10=

12−v

−12v

\maltese \: 10(12 - v) = - 12v✠10(12−v)=−12v

\maltese \: 120 - 10v = - 12v✠120−10v=−12v

\maltese \: 120 = - 12v + 10v✠120=−12v+10v

\maltese \: 120 = 2v✠120=2v

\maltese \: v = \frac{120}{2} = \cancel \frac{120}{2} ✠v=

2

120

=

2

120

\maltese \: v = 60cm✠v=60cm

Thus a screen should be placed 60cm away from the mirror to obtain a sharp image.

Now we know that for a system of mirror the ratio of size or image to size of object = Ratio of image distance to object distance. Represented by :-

After inputting the known values we get :

Therefore we get that the size of image is 25cm.

Therefore we get that the size of image is 25cm.Now we can observe that size of object < Size of image, thus the image is magnified, real and inverted.

Answer 2

Answer:

Heya,Mate,Here is Your Answer....

Step-by-step explanation:

★Given

An object 5 cm in size is placed at 12 CM in front of a concave mirror of radius of curvature of 20 CM at what distance from the mirror should a screen be placed in order to obtain a sharp image? find the nature and size of the L.

★To Find

Nature and Size of The L?

★Solution

Please Refer To Attached File

✪Formula Used=

$\tt\large\underline\orange{\longmapsto. \frac{1}{f} = \frac{1}{v} + \frac{1}{u} }$

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