Answer
64.8k+ views
Hint: In this as we know that the spherical lens is a part of a spherical mirror. The relation between focal length of a spherical lens and the radius of a spherical lens is calculated by understanding the sign convention.
Complete step by step answer:
As we know that the center of a spherical mirror or the lens is known as the center of curvature of that mirror or lens. The radius of a spherical mirror or lens is defined as the distance between the centre of curvature to the circumference of the curvature.
The figure below represents the sketch diagram of a curved surface (mirror or lens).
![](https://www.vedantu.com/question-sets/e64b17d6-6886-4699-aedc-db671409919a3870899695653807532.png)
As we can see in the above figure $CP = CB$. These both distances are the radius of the mirror.
From the geometry of the Figure it is clear that,
$ \Rightarrow BF = PF$
Further we can conclude that,
$ \Rightarrow FC = FP$
Also,
$ \Rightarrow FP = PF$
From the diagram we can write that,
$ \Rightarrow PC = PF + FC$
Now we substitute $PF$ for $FC$ in above equation.
$PC = PF + PF$
After simplification we get,
$ \Rightarrow R = 2PF$
Now substitute $f$ for$PF$ in above equation to get
$\therefore R = 2f$
Here, $f$ is the focal length of the spherical lens.
It is clear from the above expression that the radius of curvature of a spherical lens is twice the focal length of the lens.
Therefore, the correct option is (B).
Note: In this question the diagram of the spherical lens should be made clear and the basic knowledge of geometry is applied. The main concept in this question is substitution of various data (distance) from the diagram. If this substitution goes wrong the answer will be wrong.
Complete step by step answer:
As we know that the center of a spherical mirror or the lens is known as the center of curvature of that mirror or lens. The radius of a spherical mirror or lens is defined as the distance between the centre of curvature to the circumference of the curvature.
The figure below represents the sketch diagram of a curved surface (mirror or lens).
![](https://www.vedantu.com/question-sets/e64b17d6-6886-4699-aedc-db671409919a3870899695653807532.png)
As we can see in the above figure $CP = CB$. These both distances are the radius of the mirror.
From the geometry of the Figure it is clear that,
$ \Rightarrow BF = PF$
Further we can conclude that,
$ \Rightarrow FC = FP$
Also,
$ \Rightarrow FP = PF$
From the diagram we can write that,
$ \Rightarrow PC = PF + FC$
Now we substitute $PF$ for $FC$ in above equation.
$PC = PF + PF$
After simplification we get,
$ \Rightarrow R = 2PF$
Now substitute $f$ for$PF$ in above equation to get
$\therefore R = 2f$
Here, $f$ is the focal length of the spherical lens.
It is clear from the above expression that the radius of curvature of a spherical lens is twice the focal length of the lens.
Therefore, the correct option is (B).
Note: In this question the diagram of the spherical lens should be made clear and the basic knowledge of geometry is applied. The main concept in this question is substitution of various data (distance) from the diagram. If this substitution goes wrong the answer will be wrong.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)