Question

A concave lence has focal length of 15 cm .at what distance should the object from the lens. be placed so that it forms a image at 10 cm from lens n​

Answer 1

Answer:

The object is twice as far as the image, from the lens.

So u=2v

1/u+1/v=1/f

1/2v+1/v=1/10

3/2v=1/10

2v/3=10

The object is twice as far as the image, from the lens.

So u=2v

1/u+1/v=1/f

1/2v+1/v=1/10

3/2v=1/10

2v/3=10

v=15 and therefore u=30

They are 45cm apart

Hence,

option (A) is correct answer.

v=15 and therefore u=30

They are 45cm apart

Hence,

option (A) is correct answer.

Answer 2

$\large\bf\underline \blue {To \: \mathscr{f}ind:-}$

• we need to find object distance

✫$\huge\bf\underline \purple{ \mathcal{S}olution...}$

$\bf\underline{\red{Given:-}}$

• Focal length (f) = - 15cm
• Image distance (v) = - 10cm
• object distance (u) = ?

As we know that, A concave Lens always forms a virtual, erect image on the same side of the object. So , v = - 10cm

• $\large\bf\underline \red { \mathscr{L}ens\: Formula:-}$

• $\bf{\huge {\green{ \frac{1}{v} - \frac{1}{u} = \frac{1}{f}}} }$

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

$\dashrightarrow \rm {\large \frac{1}{u} = \frac{1}{ - 10} - \frac{1}{( - 15)}}$

$\large\dashrightarrow \rm {\large \frac{1}{u} = \frac{ - 1}{10} + \frac{1}{15}}$

$\large\dashrightarrow \rm {\large \frac{1}{u} = \frac{ - 3 + 2}{30}}$

$\large\dashrightarrow \rm {\large \frac{1}{u} = \frac{ - 1}{30}}$

$\dashrightarrow \bf \red{\large\:u = - 30cm} \:$

Hence,

• The object distance is -30cm

✫ Magnification = v/u

↛ m = - 10cm/- 30cm

↛ m = 10/30

↛ m = + 0.33

So,

The image is 1/3 of the size of object. And positive sign shows that the image is virtual and erect .

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