
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
133.8k+ 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

Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main Books 2023-24: Best JEE Main Books for Physics, Chemistry and Maths

JEE Main 2023 April 13 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 11 Shift 2 Question Paper with Answer Key

JEE Main 2023 April 10 Shift 2 Question Paper with Answer Key

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

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

Class 11 JEE Main Physics Mock Test 2025

JEE Main Chemistry Question Paper with Answer Keys and Solutions

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
