
The luminous intensity of a light source is 500cd. The illuminance of a surface distant 10m from it, will be if light falls normally on it
A. 5 lux
B. 10 lux
C. 20 lux
D. 40 lux
Answer
134.7k+ views
Hint: In this question, we are given the luminous intensity of a light source and the distance of the light source from the screen. We know that the illuminance of a light source is defined as the luminous intensity falling on per unit square area. We just need to substitute the given values in this equation.
Complete Step by step solution:
Luminous intensity is the amount of visible light emitted by any light source in unit time per unit solid angle. It is one of the seven fundamental physical quantities in physics. The SI unit for luminous intensity is Candela, cd.
Illuminance is the total luminous flux incident on any surface per unit area of the surface. It is measured in lux.
Illuminance, \[E = \dfrac{I}{{{R^2}}}lux\], where I is the luminous intensity and R is the distance between the surface and the light source
Given that \[I = 500cd\,and\,R = 10m\]
By substituting the values, we get that
\[ \Rightarrow E = \dfrac{{500}}{{{{10}^2}}} = 5lux\]
Note: We know that Illuminance, \[E = \dfrac{I}{{{R^2}}}lux\], is different from Luminous Intensity that is I. Illuminance is different from brightness as illuminance is the measure of intensity of light falling onto a surface, while brightness is the visual perceptions of light.
Complete Step by step solution:
Luminous intensity is the amount of visible light emitted by any light source in unit time per unit solid angle. It is one of the seven fundamental physical quantities in physics. The SI unit for luminous intensity is Candela, cd.
Illuminance is the total luminous flux incident on any surface per unit area of the surface. It is measured in lux.
Illuminance, \[E = \dfrac{I}{{{R^2}}}lux\], where I is the luminous intensity and R is the distance between the surface and the light source
Given that \[I = 500cd\,and\,R = 10m\]
By substituting the values, we get that
\[ \Rightarrow E = \dfrac{{500}}{{{{10}^2}}} = 5lux\]
Note: We know that Illuminance, \[E = \dfrac{I}{{{R^2}}}lux\], is different from Luminous Intensity that is I. Illuminance is different from brightness as illuminance is the measure of intensity of light falling onto a surface, while brightness is the visual perceptions of light.
Recently Updated Pages
JEE Main 2025 Session 2 Form Correction (Closed) – What Can Be Edited

What are examples of Chemical Properties class 10 chemistry JEE_Main

JEE Main 2025 Session 2 Schedule Released – Check Important Details Here!

JEE Main 2025 Session 2 Admit Card – Release Date & Direct Download Link

JEE Main 2025 Session 2 Registration (Closed) - Link, Last Date & Fees

JEE Mains Result 2025 NTA NIC – Check Your Score Now!

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Elastic Collisions in One Dimension - JEE Important Topic

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Motion In A Plane: Line Class 11 Notes: CBSE Physics Chapter 3
