Question

The position of the image formed will be
A 3.0 cm tall object is placed perpendicular to the principal axis of a
convex lens of focal length 10 cm. The distance of the object from the
lens is 15 cm.
0 +30 cm
0-50 cm
O +15 cm
0-10 cm​

Answer 1

Answer:

• +30 cm is the required answer

Explanation:

Given:-

• Height of object ,ho = 3 cm

• Focal length ,f = +10cm

• Object distance ,u = -15 cm

To Find:-

• image distance ,v

Solution:-

Using lens formula

• 1/v - 1/u = 1/f

where,

v is the Image Position

u is the object distance

f is the focal length

Substitute the value we get

→ 1/10 = 1/v - 1/-15

→ 1/10 = 1/v + 1/15

→ 1/v = 1/10 -1/15

→ 1/v = 3-2/30

→ 1/v = 1/30

→ v = +30 cm

• Hence, the image position is 30 cm .

Answer 2

$\\ \\ \large\underline{ \underline{ \sf{ \red{given:} }}}$

• Height of object (h) = 3cm

• Focal length ( f ) = 10cm

• Distance of object from lens (u) = - 15cm

$\\ \\ \large\underline{ \underline{ \sf{ \red{to \: find:} }}}$

• Distance of image from lens (v)

$\\ \\ \large\underline{ \underline{ \sf{ \red{solution:} }}}$

Lens formula is given by ,

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

Putting values , we get ,

$\\ \\ \sf \: \frac{1}{ 10} = \frac{1}{v} - \frac{1}{-15} \\ \\ \\ \sf \: \frac{1}{v} = \frac{1}{ 10} - \frac{1}{15} \\ \\ \\ \sf \: \frac{1}{v} = \frac{ 15 - 10}{150} \\ \\ \\ \sf \: \frac{1}{v} = \frac{ 5}{150} \\ \\ \\ \sf \: \frac{1}{v} = \frac{1}{30} \\ \\ \\ \sf \pink{ \: v = 30cm}$

Hence , position of image is 30cm behind the lens.

Opt (a) 30cm