
Minimum focal length of eye lens of normal person is
A. 25 cm
B. 1.5 cm
C. 25/9 cm
D. 25/11 cm
Answer
597.6k+ views
Hint: These are the basic problems which can be solved by using image distance (v) 2.5 cm since 2.5 cm is the distance between retina and lens then use the lens formula .i.e. $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ to find the minimum focal length.
Complete Step-by-Step solution:
Relation between u, v and f is explained by lens formula
$ \Rightarrow $$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ here u is object distance, v is image distance and f is the focal length. (Equation 1)
Since we know that distance between lens and retina is 2.5 cm, and we know that the image will form at retina therefore v = 2.5 cm
For minimum focal length:
Object distance will be 25cm (least distance of distinct vision)
Hence, u = 25 cm
Substituting the values of u and v in the equation 1
$\dfrac{1}{{2.5}} - \dfrac{1}{{ - 25}} = \dfrac{1}{{{f_{\min }}}}$
$\dfrac{{11}}{{25}} = \dfrac{1}{{{f_{\min }}}}$
${f_{\min }} = \dfrac{25}{11}$ cm
Note: We are able to see images through are eyes but everyone of us don’t know the working of eye In an eye the cornea and lens of the eye act together to form a real image on the light-sensing retina, which has its densest concentration of receptors in the fovea and a blind spot over the optic nerve and the radius of curvature of the lens of an eye is adjustable to form an image on the retina for different object distances.
Complete Step-by-Step solution:
Relation between u, v and f is explained by lens formula
$ \Rightarrow $$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ here u is object distance, v is image distance and f is the focal length. (Equation 1)
Since we know that distance between lens and retina is 2.5 cm, and we know that the image will form at retina therefore v = 2.5 cm
For minimum focal length:
Object distance will be 25cm (least distance of distinct vision)
Hence, u = 25 cm
Substituting the values of u and v in the equation 1
$\dfrac{1}{{2.5}} - \dfrac{1}{{ - 25}} = \dfrac{1}{{{f_{\min }}}}$
$\dfrac{{11}}{{25}} = \dfrac{1}{{{f_{\min }}}}$
${f_{\min }} = \dfrac{25}{11}$ cm
Note: We are able to see images through are eyes but everyone of us don’t know the working of eye In an eye the cornea and lens of the eye act together to form a real image on the light-sensing retina, which has its densest concentration of receptors in the fovea and a blind spot over the optic nerve and the radius of curvature of the lens of an eye is adjustable to form an image on the retina for different object distances.
Recently Updated Pages
What happens to glucose which enters nephron along class 10 biology CBSE

Write a dialogue with at least ten utterances between class 10 english CBSE

A circle is inscribed in an equilateral triangle and class 10 maths CBSE

When the JanmiKudian Act was passed that granted the class 10 social science CBSE

A sector containing an angle of 120 circ is cut off class 10 maths CBSE

The sum of digits of a two digit number is 13 If t-class-10-maths-ICSE

Trending doubts
The shortest day of the year in India

Write a letter to the principal requesting him to grant class 10 english CBSE

What is the median of the first 10 natural numbers class 10 maths CBSE

What is the missing number in the sequence 259142027 class 10 maths CBSE

10 examples of evaporation in daily life with explanations

State and prove the Pythagoras theorem-class-10-maths-CBSE

