Question

you are the head boy head sir of the pet clubin your school you have planned an activity at local dog pound animals shelter write a notice to the student giving details of the program


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Answer 1

Answer:

you are the head boy head sir of the pet clubin your school you have planned an activity at local dog pound animals shelter write a notice to the student giving details of the program

Answer 2

$\large\bf{\underline\color{black}{\underline\color{black}\purple{Given : }}}$

> Image distance (v) = -10cm

> Focal length (f) = -15cm

$\large\bf{\underline\color{black}{\underline\color{black}\purple{To \: Find : }}}</p><p> \:$

> Object distance (u) = ?

$\large\bf{\underline\color{black}{\underline\color{black}\purple{Solution: }}}</p><p> \: \:$

A concave lens always forms a virtual, Errect image on the same side of the object.

Since.,

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

or

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

$\large \mathrm{\frac{1}{u} = \frac{1}{-10} - \frac{1}{(-15)}} \: \\ \\ \implies \large \mathrm{\frac{1}{u} = \frac{1}{10 } + \frac{1}{15}} \: \\ \\ \implies \large \mathrm{\frac{1}{u} = \frac{ -3 + 2 } {30} } \\ \\ \implies \large \mathrm{\frac{1}{u} = \frac{1}{-30}} \\ \\ \implies \large\fbox \mathrm{u = -30cm}$

Thus the object distance is 30cm.

Magnification :

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

$\large\mathrm{ m= \frac{-10cm}{-20cm}} \\ \\ \implies\large\mathrm{ \frac{1}{3}} \\ \\ \implies \large\fbox \mathrm{ m = +0.33}$

The positive sign shows that the image is erect, virtual. The image is one-third of the size of the object.

$\huge\mathfrak\pink{Verified} \:$