
The focal length of convex lens is 30 cm and the size of the image is quarter of the object, then the object distance is:
A. 90cm
B. 60cm
C. 30cm
D. 40cm
Answer
574.2k+ views
Hint: In order to solve this problem, the relationship between the magnification of the image and the relative distances of the image and object must be determined.
For a lens,
Magnification, $m = \dfrac{v}{u}$
where v = distance of image and u = distance of the object measured from the centre of the lens.
Complete step-by-step solution:
The magnification of the image is the ratio of the size of the image to the size of the object. The magnification produced by a lens is given by the formula –
$m = \dfrac{v}{u}$
where v = distance of image and u = distance of the object measured from the centre of the lens.
Given the magnification, $m = \dfrac{1}{4}$
Applying the formula for magnification, we can obtain the relation between the distances of the object and the image.
\[\dfrac{v}{{\left( { - u} \right)}} = \dfrac{1}{4}\]
$ \Rightarrow u = - 4v$
Note that the distance of an object in any case is always negative.
The lens formula gives us the relationship between the focal length of the lens and the distances of the object and image.
Lens formula:
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$
Given, the focal length of the lens, $f = + 30cm$
Applying the focal length and the relation between v and u, we get –
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$
$\dfrac{1}{v} - \dfrac{1}{{\left( {4v} \right)}} = \dfrac{1}{{30}}$
$ \Rightarrow \dfrac{3}{{4v}} = \dfrac{1}{{30}}$
$ \Rightarrow 4v = 30 \times 3 = 90$
$ \Rightarrow v = \dfrac{{90}}{4} = 22 \cdot 5cm$
Given that, $u = - 4v$,
$u = - 4 \times 22 \cdot 5 = - 90cm$
Therefore, the object is at a distance of 90cm from the lens.
Hence, the correct option is Option A.
Note: While substituting the value of u or v in the equation, the students should take extra caution while substituting the signs. The students should follow the best practice of substituting the entire brackets with the sign. For example, if u = – 30 cm, we have to first substitute as ( – 30) in the first substitution so that there is no error in calculations.
For a lens,
Magnification, $m = \dfrac{v}{u}$
where v = distance of image and u = distance of the object measured from the centre of the lens.
Complete step-by-step solution:
The magnification of the image is the ratio of the size of the image to the size of the object. The magnification produced by a lens is given by the formula –
$m = \dfrac{v}{u}$
where v = distance of image and u = distance of the object measured from the centre of the lens.
Given the magnification, $m = \dfrac{1}{4}$
Applying the formula for magnification, we can obtain the relation between the distances of the object and the image.
\[\dfrac{v}{{\left( { - u} \right)}} = \dfrac{1}{4}\]
$ \Rightarrow u = - 4v$
Note that the distance of an object in any case is always negative.
The lens formula gives us the relationship between the focal length of the lens and the distances of the object and image.
Lens formula:
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$
Given, the focal length of the lens, $f = + 30cm$
Applying the focal length and the relation between v and u, we get –
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$
$\dfrac{1}{v} - \dfrac{1}{{\left( {4v} \right)}} = \dfrac{1}{{30}}$
$ \Rightarrow \dfrac{3}{{4v}} = \dfrac{1}{{30}}$
$ \Rightarrow 4v = 30 \times 3 = 90$
$ \Rightarrow v = \dfrac{{90}}{4} = 22 \cdot 5cm$
Given that, $u = - 4v$,
$u = - 4 \times 22 \cdot 5 = - 90cm$
Therefore, the object is at a distance of 90cm from the lens.
Hence, the correct option is Option A.
Note: While substituting the value of u or v in the equation, the students should take extra caution while substituting the signs. The students should follow the best practice of substituting the entire brackets with the sign. For example, if u = – 30 cm, we have to first substitute as ( – 30) in the first substitution so that there is no error in calculations.
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