Question

An object of height 5cm is placed Perpendicular to the principal axis of Concave lens of focal length 10cm. If the distance of the object from the optical centre of lens is 20cm. Determine the position, Nature and size of the Images formed using the lens Formula.

Answer 1

$Hey\:!!..$

The answer goes here....

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》To find -

$v$ = $?$

》Given -

$f$ = $-10\:cn$

$u$ = $-20\:cm$

》Solution -

Here, by using the formula -

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

⇒ $\frac{1}{u}$ = $\frac{1}{f}+\frac{1}{u}$

⇒ $\frac{1}{-10}+\frac{1}{-20}$  

⇒ $\frac{-3}{20}$

⇒ $v$ = $\frac{-20}{3}$

Since the image is at $\frac{-20}{3}$ on same side same as object. Therefore, the image formed is virtual.

Now, magnification -

$m$ = $\frac{h'}{h}$ = $\frac{v}{u}$

⇒ $h$ = $\frac{v}{u}h$

⇒ $\frac{-20}{3\times -20}\times 5$

⇒ $\frac{5}{3}$

⇒ $1.67\:cm$

Hence, the image formed is virtual and erect.

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Thanks !!..

Answer 2

1 - 1 = 1
v   u     f

f =- 10 cm
u = - 20 cm
so,
v = -20/3 = -6.66 cm

hi/ho = v/u
hi = -20/3 *5
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-20
    = 5/3 = 1.66 cm 
The image formed is between lens & focal length. the image is diminished, virtual & erect.