Which is the animal,that can't walk backwards?
Answers
Answer:
Kangaroos
Explanation:
Kangaroos are well known as large, hopping mammals from Australia that carry their offspring in pouches. What may not be so well known, though, is that kangaroos cannot walk backwards.
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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.
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.
