
A convex lens of focal length $1.0\,m$ and a concave lens of focal length $0.25\,m$ are $0.75\,m$ apart. A parallel beam of light is incident in the convex lens. The beam emerging after refraction from both lenses is:
A. Parallel to the principal axis
B. Convergent
C. Divergent
D. None of the above
Answer
405.6k+ views
Hint: A lens is a piece of transparent glass which is bounded by two surfaces out of which at least one surface is spherical. Refraction is the bending of light when it obliquely travels from one medium to another medium. Here the power of combination of lenses is applied.
Complete step by step answer:
From the question, convex lenses and concave lenses are $0.75{\text{ m}}$ apart.
Therefore, from the power of combination of lenses.
$\dfrac{1}{f} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}} - \dfrac{d}{{{f_1}{f_2}}}$
Were, ${f_1}$ is the focal length of convex lens $ = 1{\text{ m}}$
${f_2}$ is the focal length of concave lens $ = 0.25{\text{ m}}$
$d$ is the distance between the lenses $ = 0.75{\text{ m}}$
$\dfrac{1}{f} = \dfrac{1}{{ + 1}} + \dfrac{1}{{ - 0.25}} - \dfrac{{0.75}}{{1 \times \left( { - 0.25} \right)}}$
$ \Rightarrow \dfrac{1}{f} = 1 - 4 + 3$
Further simplifying we get,
$ \Rightarrow \dfrac{1}{f} = 4 - 4$
$ \therefore \dfrac{1}{f} = 0$
Therefore, the focal length will be infinity and the power will become zero. As the power is zero therefore if the incident beam of light is parallel then the emerging beam of light will be also parallel.
Therefore, the correct answer is option A.
Additional information:
Power of a lens: The power of a lens is defined as the reciprocal of its focal length. It is represented by the letter $P$. The power $P$ of a lens of focal length $f$ is given by,
$P = \dfrac{1}{f}$
The SI unit of power is diopter when focal length is in meters. It is noted by $D$. Hence one diopter is a power of lens whose focal length is 1 metre.
Note: A Convex lens is one which is thinner at the sides and thick at centre. A concave lens is one which is thicker at the sides and thin at the centre. When two or more lenses are combined then the power of the combined lens is the sum of individual power of lenses.
Complete step by step answer:
From the question, convex lenses and concave lenses are $0.75{\text{ m}}$ apart.
Therefore, from the power of combination of lenses.
$\dfrac{1}{f} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}} - \dfrac{d}{{{f_1}{f_2}}}$
Were, ${f_1}$ is the focal length of convex lens $ = 1{\text{ m}}$
${f_2}$ is the focal length of concave lens $ = 0.25{\text{ m}}$
$d$ is the distance between the lenses $ = 0.75{\text{ m}}$
$\dfrac{1}{f} = \dfrac{1}{{ + 1}} + \dfrac{1}{{ - 0.25}} - \dfrac{{0.75}}{{1 \times \left( { - 0.25} \right)}}$
$ \Rightarrow \dfrac{1}{f} = 1 - 4 + 3$
Further simplifying we get,
$ \Rightarrow \dfrac{1}{f} = 4 - 4$
$ \therefore \dfrac{1}{f} = 0$
Therefore, the focal length will be infinity and the power will become zero. As the power is zero therefore if the incident beam of light is parallel then the emerging beam of light will be also parallel.
Therefore, the correct answer is option A.
Additional information:
Power of a lens: The power of a lens is defined as the reciprocal of its focal length. It is represented by the letter $P$. The power $P$ of a lens of focal length $f$ is given by,
$P = \dfrac{1}{f}$
The SI unit of power is diopter when focal length is in meters. It is noted by $D$. Hence one diopter is a power of lens whose focal length is 1 metre.
Note: A Convex lens is one which is thinner at the sides and thick at centre. A concave lens is one which is thicker at the sides and thin at the centre. When two or more lenses are combined then the power of the combined lens is the sum of individual power of lenses.
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