Q. 7 : Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.
Answer:
As we know that hypermetropia is corrected by using a convex lens.
Here, The near point of a hypermetropic eye is 1 m and the near point of the normal eye is 25 cm.
Therefor, v= − 1 metre = − 100 cm and u= −25 cm
As per lens formula :
1  1  1  

− 



v  u  f  
1  1  1  

− 



− 100  − 25  f  
− 1 + 4  1  

− 


100  f  
100  
f  = 

cm  
3  
1  
f  = 

m  
3  
1  
P  = 


f  
1 × 3  
P  = 

Dioptre  
1  
P  =  + 3.0  D 