Question

A convex lens of focal length 6cm is held 4 cm from a newspaper which has print 0.5cm high. By calculation , determine the size and nature of the image formed.

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Answer 1

Given:
Focal length, f = 6 cm
Object distance, u = -4 cm
Size of the object, h = 0.5 cm
Applying lens formula, we get:
1/v -1/u = 1/f
or, 1/v = 1/f + 1/u
= 1/6 + 1/(-4) = -1/12
or, image distance, v = -12 cm
Applying magnification formula, we get:
m = v/u = h'/h

or, m = (-12)/(-4) = h'/0.5

or,  3 = h'/0.5
or,      h, =3×0.5
Size of the image, h' = 1.5 cm
Therefore, the height of the image is 1.5 cm. The value of magnification is positive; therefore, the image is virtual, erect and three times magnified.

Answer 2

heya !!

here is ur answer
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Given data :-

focal length of the convex lens = 6 cm

distance from object to lens (u ) = 4 cm

height of object ( hò) = 0.5 cm

height of image = ??

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to find height of image v need to find image distance .

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lens formula :-

$\frac{1}{v} - \frac{1}{u } \: = \frac{1}{f}$

1/v -(- 1/4) = 1/6

1/v + 1/4 = 1/6

1/v = 1/6- 1/4

LCM = 12

1/v = 2-3/12

1/v = -1 / 12

-v = 12

v = -12

we got "v" as -12

to find the height of image v use the formula :-

$\frac{hi}{ho} = \frac{( - v)}{u}$

↪hì / 0.5 = -(-12) /4

↪ hì /0.5 = 12/4

↪ hì / 0.5 = 3/1

↪ hì = 0.5 × 3

↪ hì = 1.5 cm

↪ so here the height of image is more than height of object

↪ the object is kept between focus and pole .

♦ nature of image formed :-

↪ virtual

↪ upright

↪ magnifide

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hope it helps u ..!!