
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
216.3k+ 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 Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Electricity and Magnetism Explained: Key Concepts & Applications

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

NCERT Solutions For Class 11 Physics Chapter 8 Mechanical Properties Of Solids

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26

NCERT Solutions for Class 11 Physics Chapter 7 Gravitation 2025-26

Understanding Atomic Structure for Beginners

