Question

$\begin{gathered}{\Huge{\textsf{\textbf{\underline{\underline{\color{red}{Question:}}}}}}}\end{gathered}$
$\textsf{If an object of 7cm height is placed at }\\ \textsf{a distance of 12cm from a convex} \\ \textsf{lens of focal length 8cm, find the} \\ \textsf{ position, nature and height of the image.}$
$\large \bf \underline\red{Note}$
$\star{\textsf{Need Full Explanation}}$
$\star \textsf{Don't Spam}$
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Answer 1

Explanation:

First of all we find out the position of image by the position of the image beam in the distance of image from The Lens HERE object distance , u= - 12 cm (it is too left lens ) image distance, v = ? ( to be calculated ) focal length , f = +8 cm (it is convex lens) putting this value in the lens formula ;

.

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

$\frac{1}{v } - \frac{1}{ - 12} = \frac{1}{8}$

$\frac{1}{v} + \frac{1}{12} = \frac{1}{8}$

$\frac{1}{v} = \frac{1}{8} - \frac{1}{12} \\ \frac{1}{v } = \frac{3 - 2}{24} \\ \frac{1}{v} = + 24$

so image distance V = +24

Thus the image is formed on the right side of the convex lens only a real and inverted image is formed on the right side of convex lens show the image formed is real and inverted let's Earth calculate the magnification now we know that for lens magnification.

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

here image distance V = 24 object distance U.

$- 12$

$m = \frac{24}{ - 12}$

$m = - 2$

The value of magnification is more than 1 so the image is larger than the object of magnified the minus sign for magnification show the that the image is formed below the principal Axis hence the image is real and inverted let are calculated the size of the image by using formula.

$m = \frac{h2}{h1}$

$m = - 2$

height of object

$h1 = - 2$

height of image

$h1 = + 7$ height of image

$h2 = find$

now putting this value in the above formula .

$- 2 = \frac{h2}{7} \\ h2 = - 2 \times 7 \\ height \: of \: image \\ h2 = - 14$

the height of the size of height image is 14 cm and the minus sign so that the height is in the downward direction that the direction form below the axis the image is real and inverted .

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