Question

answer the question in attachment

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Answer 1

Solution

In general, to find object distance, image distance, or focal length of a concave/convex lens, we use the Lens formula:

$\tt{\Longrightarrow \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}}$

In the above formula,

$\tt{\implies v = Image distance = -10\:cm}$

$\tt{\implies f = Focal length = -15\:cm}$

$\tt{\implies u = Object distance}$

Note that in a concave lens, the values of image distance, object distance, and focal length are always negative.

According to the question, we need to find the object distance.

By applying the Lens formula, we get:

$\tt{\Longrightarrow \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}}$

$\tt{\Longrightarrow \dfrac{-1}{10} - \dfrac{1}{u} = \dfrac{-1}{15}}$

$\tt{\Longrightarrow \dfrac{-1}{u} = \dfrac{-1}{15} + \dfrac{1}{10}}$

$\tt{\Longrightarrow \dfrac{-1}{u} = \dfrac{-2+3}{30}}$

$\tt{\Longrightarrow \dfrac{-1}{u} = \dfrac{1}{30}}$

$\tt{\Longrightarrow u = -30\:cm}$

Hence, the object is placed 30 cm away from the concave lens.

$\rule{310}{2}$

To find magnification of the lens, we use the formula:

$\tt{\Longrightarrow m = \dfrac{v}{u}}$

$\tt{\Longrightarrow m = \dfrac{-10}{-30}}$

$\tt{\Longrightarrow m = \dfrac{1}{3} \: (or) \: 0.3}$

$\rule{310}{2}$

The image formed is always virtual, erect, and diminished in size for concave lens.

Answer 2

Answer:

Solution

In general, to find object distance, image distance, or focal length of a concave/convex lens, we use the Lens formula:

\tt{\Longrightarrow \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}}⟹

v

1

−

u

1

=

f

1

In the above formula,

\tt{\implies v = Image distance = -10\:cm}⟹v=Imagedistance=−10cm

\tt{\implies f = Focal length = -15\:cm}⟹f=Focallength=−15cm

\tt{\implies u = Object distance}⟹u=Objectdistance

Note that in a concave lens, the values of image distance, object distance, and focal length are always negative.

According to the question, we need to find the object distance.

By applying the Lens formula, we get:

\tt{\Longrightarrow \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}}⟹

v

1

−

u

1

=

f

1

\tt{\Longrightarrow \dfrac{-1}{10} - \dfrac{1}{u} = \dfrac{-1}{15}}⟹

10

−1

−

u

1

=

15

−1

\tt{\Longrightarrow \dfrac{-1}{u} = \dfrac{-1}{15} + \dfrac{1}{10}}⟹

u

−1

=

15

−1

+

10

1

\tt{\Longrightarrow \dfrac{-1}{u} = \dfrac{-2+3}{30}}⟹

u

−1

=

30

−2+3

\tt{\Longrightarrow \dfrac{-1}{u} = \dfrac{1}{30}}⟹

u

−1

=

30

1

\tt{\Longrightarrow u = -30\:cm}⟹u=−30cm

Hence, the object is placed 30 cm away from the concave lens.

\rule{310}{2}

To find magnification of the lens, we use the formula:

\tt{\Longrightarrow m = \dfrac{v}{u}}⟹m=

u

v

\tt{\Longrightarrow m = \dfrac{-10}{-30}}⟹m=

−30

−10

\tt{\Longrightarrow m = \dfrac{1}{3} \: (or) \: 0.3}⟹m=

3

1

(or)0.3

\rule{310}{2}

The image formed is always virtual, erect, and diminished in size for concave lens.