
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
596.4k+ 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.
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