AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

[28]
A person cannot see objects kept at $2 m$ distinctly. The defect can be corrected by using a lens of the Power _________.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

+ 0.5 D

+ 0.2 D

- 0.5 D

- 0.2 D

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

A person cannot see the objects kept beyond

2 m

distinctly. This defect can be corrected by using a lens of power

- 0.5 D

.
The far point of a typical eye is infinity. The defective eye's far point is given as

2 m

. The eye is short-sighted. To correct this defect, one should bring the object from infinity to

200 cm

. (as

1 m = 100 cm

)
From the data given,

u = - \infty

v = - 200 cm

f = ?

Substituting in lens formula,

\frac{1}{f} = \frac{1}{v} - \frac{1}{u}

where

u =

the distance between the object and the pole of the mirror,

v =

the distance between the image and the pole of the mirror,

f =

focal length
We get,

\frac{1}{f} = \frac{1}{(- 200)} - \frac{1}{(- \infty)}

\Rightarrow \frac{1}{f} = \frac{1}{(- 200)} + 0

(as

1 / \infty = 0

)

\Rightarrow f = - 200 cm

We know that

Power of the lens = \frac{100}{f (in cm)}

\Rightarrow P = \frac{100}{- 200}

\Rightarrow P = - 0.5 D

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!